GSEB Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Exercise 5.5

Question 1. \( \cos x \cdot \cos 2x \cdot \cos 3x \)
Answer: Let \( y = \cos x \cdot \cos 2x \cdot \cos 3x \).
Taking the logarithm on both sides, we obtain:
\( \log y = \log (\cos x \cdot \cos 2x \cdot \cos 3x) \)
Using the logarithm property \( \log(abc) = \log a + \log b + \log c \), we get:
\( \log y = \log \cos x + \log \cos 2x + \log \cos 3x \)
Now, differentiating both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (\log \cos x) + \frac{d}{dx} (\log \cos 2x) + \frac{d}{dx} (\log \cos 3x) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} (-\sin x) + \frac{1}{\cos 2x} (-\sin 2x) \cdot 2 + \frac{1}{\cos 3x} (-\sin 3x) \cdot 3 \)
\( \frac{1}{y} \frac{dy}{dx} = -\frac{\sin x}{\cos x} - \frac{2 \sin 2x}{\cos 2x} - \frac{3 \sin 3x}{\cos 3x} \)
\( \frac{1}{y} \frac{dy}{dx} = -\tan x - 2\tan 2x - 3\tan 3x \)
\( \frac{1}{y} \frac{dy}{dx} = -(\tan x + 2\tan 2x + 3\tan 3x) \)
Finally, multiply by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = -y (\tan x + 2\tan 2x + 3\tan 3x) \)
Substitute back \( y = \cos x \cdot \cos 2x \cdot \cos 3x \):
\( \frac{dy}{dx} = -\cos x \cdot \cos 2x \cdot \cos 3x (\tan x + 2\tan 2x + 3\tan 3x) \)
In simple words: We find the derivative by first taking the logarithm of both sides. This helps to break the product into a sum. Then, we differentiate each term separately and combine them to get the final answer.

Exam Tip: Remember to apply the chain rule correctly when differentiating terms like \( \log \cos(nx) \), ensuring you multiply by the derivative of the inner function.

Question 2. \( \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \)
Answer: Let \( y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \).
We can write this as:
\( y = \left( \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right)^{\frac{1}{2}} \)
Taking the natural logarithm of both sides, we get:
\( \log y = \log \left( \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right)^{\frac{1}{2}} \)
Using the logarithm property \( \log a^b = b \log a \):
\( \log y = \frac{1}{2} \log \left( \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right) \)
Using the logarithm property \( \log \frac{A}{B} = \log A - \log B \):
\( \log y = \frac{1}{2} [\log ((x-1)(x-2)) - \log ((x-3)(x-4)(x-5))] \)
Using the logarithm property \( \log(AB) = \log A + \log B \):
\( \log y = \frac{1}{2} [\log (x-1) + \log (x-2) - (\log (x-3) + \log (x-4) + \log (x-5))] \)
\( \log y = \frac{1}{2} [\log (x-1) + \log (x-2) - \log (x-3) - \log (x-4) - \log (x-5)] \)
Now, differentiating both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx}(\log (x-1)) + \frac{d}{dx}(\log (x-2)) - \frac{d}{dx}(\log (x-3)) - \frac{d}{dx}(\log (x-4)) - \frac{d}{dx}(\log (x-5)) \right] \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] \)
Multiplying by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \cdot \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] \)
Substitute back \( y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \):
\( \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] \)
In simple words: For this type of complex fraction under a square root, taking logarithms helps a lot. It simplifies the expression into a sum and difference of simpler log terms, making differentiation easier.

Exam Tip: Always remember to convert roots to fractional powers before taking logarithms. The chain rule should be correctly applied to each term in the sum.

Question 3. \( (\log x)^{\cos x} \)
Answer: Let \( y = (\log x)^{\cos x} \).
Since both the base and the exponent are functions of \( x \), we use logarithmic differentiation.
Taking the natural logarithm of both sides, we get:
\( \log y = \log ((\log x)^{\cos x}) \)
Using the logarithm property \( \log a^b = b \log a \):
\( \log y = \cos x \cdot \log (\log x) \)
Now, differentiate both sides with respect to \( x \), using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (\cos x) \cdot \log (\log x) + \cos x \cdot \frac{d}{dx} (\log (\log x)) \)
For the first term, \( \frac{d}{dx} (\cos x) = -\sin x \).
For the second term, using the chain rule:
\( \frac{d}{dx} (\log (\log x)) = \frac{1}{\log x} \cdot \frac{d}{dx} (\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \)
Substitute these derivatives back into the equation:
\( \frac{1}{y} \frac{dy}{dx} = -\sin x \log (\log x) + \cos x \cdot \frac{1}{x \log x} \)
\( \frac{1}{y} \frac{dy}{dx} = -\sin x \log (\log x) + \frac{\cos x}{x \log x} \)
Multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left( -\sin x \log (\log x) + \frac{\cos x}{x \log x} \right) \)
Substitute back \( y = (\log x)^{\cos x} \):
\( \frac{dy}{dx} = (\log x)^{\cos x} \left( -\sin x \log (\log x) + \frac{\cos x}{x \log x} \right) \)
In simple words: When you have a function where both the base and the power are variables, use logarithms to bring the power down. Then, apply the product rule and chain rule carefully to find the derivative.

Exam Tip: Be careful with the chain rule for \( \log (\log x) \). It involves differentiating \( \log u \) where \( u = \log x \), leading to \( \frac{1}{u} \cdot \frac{du}{dx} \).

Question 4. \( x^x - 2^{\sin x} \)
Answer: Let \( y = x^x - 2^{\sin x} \).
This expression has two terms, each requiring logarithmic differentiation or specific rules. Let \( u = x^x \) and \( v = 2^{\sin x} \).
So, \( y = u - v \), which means \( \frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx} \).

First, consider \( u = x^x \).
Taking the natural logarithm of both sides: \( \log u = \log (x^x) \)
Using \( \log a^b = b \log a \): \( \log u = x \log x \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x) \)
\( \frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} \)
\( \frac{1}{u} \frac{du}{dx} = \log x + 1 \)
Multiply by \( u \): \( \frac{du}{dx} = u(1 + \log x) \)
Substitute back \( u = x^x \): \( \frac{du}{dx} = x^x (1 + \log x) \).

Next, consider \( v = 2^{\sin x} \).
This is of the form \( a^f(x) \), whose derivative is \( a^{f(x)} \log a \cdot f'(x) \).
Here, \( a = 2 \) and \( f(x) = \sin x \). So \( f'(x) = \cos x \).
Therefore, \( \frac{dv}{dx} = 2^{\sin x} \log 2 \cdot \cos x \).

Finally, combine \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\( \frac{dy}{dx} = x^x (1 + \log x) - 2^{\sin x} \log 2 \cdot \cos x \)
In simple words: This problem involves two different types of functions. We find the derivative of each part separately and then subtract them. One part needs logarithmic differentiation, while the other uses a standard exponential function rule.

Exam Tip: For expressions like \( x^x \), always use logarithmic differentiation. For expressions like \( a^{f(x)} \), remember the rule \( \frac{d}{dx}(a^{f(x)}) = a^{f(x)} \log a \cdot f'(x) \).

Question 5. \( (x + 3)^2 \cdot (x + 4)^3 \cdot (x + 5)^4 \)
Answer: Let \( y = (x + 3)^2 \cdot (x + 4)^3 \cdot (x + 5)^4 \).
Since this is a product of multiple functions raised to powers, logarithmic differentiation simplifies the process.
Taking the natural logarithm of both sides, we get:
\( \log y = \log ((x + 3)^2 \cdot (x + 4)^3 \cdot (x + 5)^4) \)
Using the logarithm property \( \log(abc) = \log a + \log b + \log c \):
\( \log y = \log (x + 3)^2 + \log (x + 4)^3 + \log (x + 5)^4 \)
Using the logarithm property \( \log a^b = b \log a \):
\( \log y = 2 \log (x + 3) + 3 \log (x + 4) + 4 \log (x + 5) \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (2 \log (x + 3)) + \frac{d}{dx} (3 \log (x + 4)) + \frac{d}{dx} (4 \log (x + 5)) \)
\( \frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{1}{x + 3} \cdot 1 + 3 \cdot \frac{1}{x + 4} \cdot 1 + 4 \cdot \frac{1}{x + 5} \cdot 1 \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5} \)
Multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left( \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5} \right) \)
Substitute back \( y = (x + 3)^2 \cdot (x + 4)^3 \cdot (x + 5)^4 \):
\( \frac{dy}{dx} = (x + 3)^2 \cdot (x + 4)^3 \cdot (x + 5)^4 \left( \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5} \right) \)
In simple words: When you have a product of many terms, each raised to a power, using logarithms helps simplify it into a sum. Differentiating a sum is much simpler than using the product rule multiple times.

Exam Tip: Logarithmic differentiation is very useful for functions involving products, quotients, or powers of multiple expressions. Make sure to apply the log properties correctly before differentiating.

Question 6. \( \left(x+\frac{1}{x}\right)^{x} + x^{\left(1+\frac{1}{x}\right)} \)
Answer: Let \( y = \left(x+\frac{1}{x}\right)^{x} + x^{\left(1+\frac{1}{x}\right)} \).
Let \( u = \left(x+\frac{1}{x}\right)^{x} \) and \( v = x^{\left(1+\frac{1}{x}\right)} \).
So, \( y = u + v \), which means \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).

First, find \( \frac{du}{dx} \) for \( u = \left(x+\frac{1}{x}\right)^{x} \).
Taking the natural logarithm of both sides: \( \log u = \log \left(x+\frac{1}{x}\right)^{x} \)
Using \( \log a^b = b \log a \): \( \log u = x \log \left(x+\frac{1}{x}\right) \)
\( \log u = x \log \left(\frac{x^2+1}{x}\right) \)
\( \log u = x [\log (x^2+1) - \log x] \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) [\log (x^2+1) - \log x] + x \frac{d}{dx}[\log (x^2+1) - \log x] \)
\( \frac{1}{u} \frac{du}{dx} = 1 \cdot [\log (x^2+1) - \log x] + x \left[ \frac{1}{x^2+1} \cdot 2x - \frac{1}{x} \right] \)
\( \frac{1}{u} \frac{du}{dx} = \log \left(\frac{x^2+1}{x}\right) + x \left[ \frac{2x}{x^2+1} - \frac{1}{x} \right] \)
\( \frac{1}{u} \frac{du}{dx} = \log \left(x+\frac{1}{x}\right) + x \left[ \frac{2x^2 - (x^2+1)}{x(x^2+1)} \right] \)
\( \frac{1}{u} \frac{du}{dx} = \log \left(x+\frac{1}{x}\right) + x \left[ \frac{x^2 - 1}{x(x^2+1)} \right] \)
\( \frac{1}{u} \frac{du}{dx} = \log \left(x+\frac{1}{x}\right) + \frac{x^2 - 1}{x^2+1} \)
\( \frac{du}{dx} = u \left( \log \left(x+\frac{1}{x}\right) + \frac{x^2 - 1}{x^2+1} \right) \)
\( \frac{du}{dx} = \left(x+\frac{1}{x}\right)^{x} \left( \log \left(x+\frac{1}{x}\right) + \frac{x^2 - 1}{x^2+1} \right) \).

Next, find \( \frac{dv}{dx} \) for \( v = x^{\left(1+\frac{1}{x}\right)} \).
Taking the natural logarithm of both sides: \( \log v = \log \left(x^{\left(1+\frac{1}{x}\right)}\right) \)
Using \( \log a^b = b \log a \): \( \log v = \left(1+\frac{1}{x}\right) \log x \)
\( \log v = \left(\frac{x+1}{x}\right) \log x \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}\left(\frac{x+1}{x}\right) \log x + \left(\frac{x+1}{x}\right) \frac{d}{dx}(\log x) \)
For \( \frac{d}{dx}\left(\frac{x+1}{x}\right) \), use the quotient rule: \( \frac{1 \cdot x - (x+1) \cdot 1}{x^2} = \frac{x - x - 1}{x^2} = -\frac{1}{x^2} \).
So,
\( \frac{1}{v} \frac{dv}{dx} = -\frac{1}{x^2} \log x + \left(\frac{x+1}{x}\right) \frac{1}{x} \)
\( \frac{1}{v} \frac{dv}{dx} = -\frac{\log x}{x^2} + \frac{x+1}{x^2} \)
\( \frac{1}{v} \frac{dv}{dx} = \frac{x+1 - \log x}{x^2} \)
Multiply by \( v \): \( \frac{dv}{dx} = v \frac{x+1 - \log x}{x^2} \)
Substitute back \( v = x^{\left(1+\frac{1}{x}\right)} \): \( \frac{dv}{dx} = x^{\left(1+\frac{1}{x}\right)} \frac{x+1 - \log x}{x^2} \).

Finally, combine \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\( \frac{dy}{dx} = \left(x+\frac{1}{x}\right)^{x} \left( \log \left(x+\frac{1}{x}\right) + \frac{x^2 - 1}{x^2+1} \right) + x^{\left(1+\frac{1}{x}\right)} \frac{x+1 - \log x}{x^2} \)
In simple words: This function is a sum of two complex terms, both with variables in their base and exponent. We differentiate each term separately using logarithmic differentiation and then add the results together.

Exam Tip: When dealing with sums of such terms, always separate them into \( u \) and \( v \), differentiate each independently, and then add their derivatives. This helps prevent errors and keeps the work clear.

Question 7. \( (\log x)^x + x^{\log x} \)
Answer: Let \( y = (\log x)^x + x^{\log x} \).
Let \( u = (\log x)^x \) and \( v = x^{\log x} \).
So, \( y = u + v \), which means \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).

First, find \( \frac{du}{dx} \) for \( u = (\log x)^x \).
Taking the natural logarithm of both sides: \( \log u = \log ((\log x)^x) \)
Using \( \log a^b = b \log a \): \( \log u = x \log (\log x) \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log (\log x) + x \cdot \frac{d}{dx}(\log (\log x)) \)
\( \frac{1}{u} \frac{du}{dx} = 1 \cdot \log (\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} \)
\( \frac{1}{u} \frac{du}{dx} = \log (\log x) + \frac{1}{\log x} \)
Multiply by \( u \): \( \frac{du}{dx} = u \left( \log (\log x) + \frac{1}{\log x} \right) \)
Substitute back \( u = (\log x)^x \): \( \frac{du}{dx} = (\log x)^x \left( \log (\log x) + \frac{1}{\log x} \right) \).

Next, find \( \frac{dv}{dx} \) for \( v = x^{\log x} \).
Taking the natural logarithm of both sides: \( \log v = \log (x^{\log x}) \)
Using \( \log a^b = b \log a \): \( \log v = (\log x) \cdot (\log x) = (\log x)^2 \)
Differentiating both sides with respect to \( x \) using the chain rule:
\( \frac{1}{v} \frac{dv}{dx} = 2 (\log x) \cdot \frac{d}{dx}(\log x) \)
\( \frac{1}{v} \frac{dv}{dx} = 2 (\log x) \cdot \frac{1}{x} \)
\( \frac{1}{v} \frac{dv}{dx} = \frac{2 \log x}{x} \)
Multiply by \( v \): \( \frac{dv}{dx} = v \frac{2 \log x}{x} \)
Substitute back \( v = x^{\log x} \): \( \frac{dv}{dx} = x^{\log x} \frac{2 \log x}{x} \).

Finally, combine \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\( \frac{dy}{dx} = (\log x)^x \left( \log (\log x) + \frac{1}{\log x} \right) + x^{\log x} \frac{2 \log x}{x} \)
In simple words: We treat the two parts of the function separately, finding the derivative of each using logarithms. After calculating each derivative, we add them to get the total derivative.

Exam Tip: Remember that \( (\log x)^2 \) is differentiated using the chain rule \( (f(x))^n \implies n(f(x))^{n-1} f'(x) \), where \( f(x) = \log x \).

Question 8. \( (\sin x)^x + \sin^{-1}\sqrt{x} \)
Answer: Let \( y = (\sin x)^x + \sin^{-1}\sqrt{x} \).
Let \( u = (\sin x)^x \) and \( v = \sin^{-1}\sqrt{x} \).
So, \( y = u + v \), which means \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).

First, find \( \frac{du}{dx} \) for \( u = (\sin x)^x \).
Taking the natural logarithm of both sides: \( \log u = \log ((\sin x)^x) \)
Using \( \log a^b = b \log a \): \( \log u = x \log (\sin x) \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log (\sin x) + x \cdot \frac{d}{dx}(\log (\sin x)) \)
\( \frac{1}{u} \frac{du}{dx} = 1 \cdot \log (\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x \)
\( \frac{1}{u} \frac{du}{dx} = \log (\sin x) + x \cot x \)
Multiply by \( u \): \( \frac{du}{dx} = u (\log (\sin x) + x \cot x) \)
Substitute back \( u = (\sin x)^x \): \( \frac{du}{dx} = (\sin x)^x (\log (\sin x) + x \cot x) \).

Next, find \( \frac{dv}{dx} \) for \( v = \sin^{-1}\sqrt{x} \).
Using the chain rule, \( \frac{d}{dx}(\sin^{-1}f(x)) = \frac{1}{\sqrt{1 - (f(x))^2}} \cdot f'(x) \).
Here, \( f(x) = \sqrt{x} \), so \( f'(x) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \).
Therefore,
\( \frac{dv}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}} \)
\( \frac{dv}{dx} = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} \)
\( \frac{dv}{dx} = \frac{1}{2\sqrt{x(1 - x)}} = \frac{1}{2\sqrt{x - x^2}} \).

Finally, combine \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\( \frac{dy}{dx} = (\sin x)^x (\log (\sin x) + x \cot x) + \frac{1}{2\sqrt{x - x^2}} \)
In simple words: We have two terms to differentiate. For the first term, we use logarithms. For the second term, we use the chain rule for inverse sine functions. We then add the results from both differentiations.

Exam Tip: Remember the standard derivative for inverse trigonometric functions. Specifically, \( \frac{d}{dx}(\sin^{-1} u) = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \). Don't forget the derivative of \( \sqrt{x} \).

Question 9. \( x^{\sin x} + (\sin x)^{\cos x} \)
Answer: Let \( y = x^{\sin x} + (\sin x)^{\cos x} \).
Let \( u = x^{\sin x} \) and \( v = (\sin x)^{\cos x} \).
So, \( y = u + v \), which means \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).

First, find \( \frac{du}{dx} \) for \( u = x^{\sin x} \).
Taking the natural logarithm of both sides: \( \log u = \log (x^{\sin x}) \)
Using \( \log a^b = b \log a \): \( \log u = \sin x \log x \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(\sin x) \cdot \log x + \sin x \cdot \frac{d}{dx}(\log x) \)
\( \frac{1}{u} \frac{du}{dx} = \cos x \cdot \log x + \sin x \cdot \frac{1}{x} \)
\( \frac{1}{u} \frac{du}{dx} = \cos x \log x + \frac{\sin x}{x} \)
Multiply by \( u \): \( \frac{du}{dx} = u \left( \cos x \log x + \frac{\sin x}{x} \right) \)
Substitute back \( u = x^{\sin x} \): \( \frac{du}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right) \).

Next, find \( \frac{dv}{dx} \) for \( v = (\sin x)^{\cos x} \).
Taking the natural logarithm of both sides: \( \log v = \log ((\sin x)^{\cos x}) \)
Using \( \log a^b = b \log a \): \( \log v = \cos x \log (\sin x) \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(\cos x) \cdot \log (\sin x) + \cos x \cdot \frac{d}{dx}(\log (\sin x)) \)
\( \frac{1}{v} \frac{dv}{dx} = (-\sin x) \log (\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x \)
\( \frac{1}{v} \frac{dv}{dx} = -\sin x \log (\sin x) + \frac{\cos^2 x}{\sin x} \)
\( \frac{1}{v} \frac{dv}{dx} = -\sin x \log (\sin x) + \cos x \cot x \)
Multiply by \( v \): \( \frac{dv}{dx} = v \left( -\sin x \log (\sin x) + \cos x \cot x \right) \)
Substitute back \( v = (\sin x)^{\cos x} \): \( \frac{dv}{dx} = (\sin x)^{\cos x} \left( -\sin x \log (\sin x) + \frac{\cos^2 x}{\sin x} \right) \)
\( \frac{dv}{dx} = (\sin x)^{\cos x} \left( -\sin x \log (\sin x) + \frac{\cos^2 x - \sin^2 x \log (\sin x)}{\sin x} \right) \).

Finally, combine \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\( \frac{dy}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} \left( \frac{\cos^2 x}{\sin x} - \sin x \log (\sin x) \right) \)
In simple words: We find the derivative of each part of the sum separately using the logarithm trick, since both parts have variables in their base and exponent. Then, we add the two derivatives together.

Exam Tip: Pay close attention to the product rule and chain rule applications for both \( u \) and \( v \), especially for terms like \( \log(\sin x) \).

Question 10. \( x^{x \cos x} + \frac{x^{2}+1}{x^{2}-1} \)
Answer: Let \( y = x^{x \cos x} + \frac{x^{2}+1}{x^{2}-1} \).
Let \( u = x^{x \cos x} \) and \( v = \frac{x^{2}+1}{x^{2}-1} \).
So, \( y = u + v \), which means \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).

First, find \( \frac{du}{dx} \) for \( u = x^{x \cos x} \).
Taking the natural logarithm of both sides: \( \log u = \log (x^{x \cos x}) \)
Using \( \log a^b = b \log a \): \( \log u = (x \cos x) \log x \)
Differentiating both sides with respect to \( x \) using the product rule (three terms for the derivative of \( x \cos x \log x \)):
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \cos x \cdot \log x + x \cdot \frac{d}{dx}(\cos x) \cdot \log x + x \cdot \cos x \cdot \frac{d}{dx}(\log x) \)
\( \frac{1}{u} \frac{du}{dx} = 1 \cdot \cos x \log x + x \cdot (-\sin x) \cdot \log x + x \cdot \cos x \cdot \frac{1}{x} \)
\( \frac{1}{u} \frac{du}{dx} = \cos x \log x - x \sin x \log x + \cos x \)
Multiply by \( u \): \( \frac{du}{dx} = u (\cos x \log x - x \sin x \log x + \cos x) \)
Substitute back \( u = x^{x \cos x} \): \( \frac{du}{dx} = x^{x \cos x} (\cos x \log x - x \sin x \log x + \cos x) \).

Next, find \( \frac{dv}{dx} \) for \( v = \frac{x^{2}+1}{x^{2}-1} \).
Using the quotient rule: \( \frac{d}{dx}\left(\frac{P}{Q}\right) = \frac{P'Q - PQ'}{Q^2} \).
Here, \( P = x^2+1 \) and \( Q = x^2-1 \).
\( P' = 2x \) and \( Q' = 2x \).
Therefore,
\( \frac{dv}{dx} = \frac{(2x)(x^2-1) - (x^2+1)(2x)}{(x^2-1)^2} \)
\( \frac{dv}{dx} = \frac{2x^3 - 2x - (2x^3 + 2x)}{(x^2-1)^2} \)
\( \frac{dv}{dx} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2-1)^2} \)
\( \frac{dv}{dx} = \frac{-4x}{(x^2-1)^2} \).

Finally, combine \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\( \frac{dy}{dx} = x^{x \cos x} (\cos x \log x - x \sin x \log x + \cos x) + \frac{-4x}{(x^2-1)^2} \)
In simple words: We split the function into two parts. The first part needs logarithmic differentiation because of its complex power. The second part is a fraction, so we use the quotient rule. Then, we add the derivatives of both parts.

Exam Tip: When differentiating \( x \cos x \log x \), remember it's a product of three functions. You can either group them as \( (x \cos x) \log x \) or apply the triple product rule for \( (fgh)' = f'gh + fg'h + fgh' \).

Question 11. \( (x \cos x)^x + (x \sin x)^{\frac{1}{x}} \)
Answer: Let \( y = (x \cos x)^x + (x \sin x)^{\frac{1}{x}} \).
Let \( u = (x \cos x)^x \) and \( v = (x \sin x)^{\frac{1}{x}} \).
So, \( y = u + v \), which means \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).

First, find \( \frac{du}{dx} \) for \( u = (x \cos x)^x \).
Taking the natural logarithm of both sides: \( \log u = \log ((x \cos x)^x) \)
Using \( \log a^b = b \log a \): \( \log u = x \log (x \cos x) \)
Using \( \log(AB) = \log A + \log B \): \( \log u = x (\log x + \log \cos x) \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) (\log x + \log \cos x) + x \frac{d}{dx}(\log x + \log \cos x) \)
\( \frac{1}{u} \frac{du}{dx} = 1 \cdot (\log x + \log \cos x) + x \left( \frac{1}{x} + \frac{1}{\cos x} (-\sin x) \right) \)
\( \frac{1}{u} \frac{du}{dx} = \log x + \log \cos x + 1 - \frac{x \sin x}{\cos x} \)
\( \frac{1}{u} \frac{du}{dx} = \log (x \cos x) + 1 - x \tan x \)
Multiply by \( u \): \( \frac{du}{dx} = u (\log (x \cos x) + 1 - x \tan x) \)
Substitute back \( u = (x \cos x)^x \): \( \frac{du}{dx} = (x \cos x)^x (1 + \log (x \cos x) - x \tan x) \).

Next, find \( \frac{dv}{dx} \) for \( v = (x \sin x)^{\frac{1}{x}} \).
Taking the natural logarithm of both sides: \( \log v = \log ((x \sin x)^{\frac{1}{x}}) \)
Using \( \log a^b = b \log a \): \( \log v = \frac{1}{x} \log (x \sin x) \)
Using \( \log(AB) = \log A + \log B \): \( \log v = \frac{1}{x} (\log x + \log \sin x) \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}\left(\frac{1}{x}\right) (\log x + \log \sin x) + \frac{1}{x} \frac{d}{dx}(\log x + \log \sin x) \)
\( \frac{1}{v} \frac{dv}{dx} = \left(-\frac{1}{x^2}\right) (\log x + \log \sin x) + \frac{1}{x} \left( \frac{1}{x} + \frac{1}{\sin x} \cos x \right) \)
\( \frac{1}{v} \frac{dv}{dx} = -\frac{\log (x \sin x)}{x^2} + \frac{1}{x^2} + \frac{\cos x}{x \sin x} \)
\( \frac{1}{v} \frac{dv}{dx} = \frac{-\log (x \sin x) + 1 + x \cot x}{x^2} \)
Multiply by \( v \): \( \frac{dv}{dx} = v \frac{1 + x \cot x - \log (x \sin x)}{x^2} \)
Substitute back \( v = (x \sin x)^{\frac{1}{x}} \): \( \frac{dv}{dx} = (x \sin x)^{\frac{1}{x}} \frac{1 + x \cot x - \log (x \sin x)}{x^2} \).

Finally, combine \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\( \frac{dy}{dx} = (x \cos x)^x (1 + \log (x \cos x) - x \tan x) + (x \sin x)^{\frac{1}{x}} \frac{1 + x \cot x - \log (x \sin x)}{x^2} \)
In simple words: This problem involves two complicated terms added together. We use logarithmic differentiation for each term, applying the product rule and chain rule carefully. Finally, we add the derivatives of the two terms.

Exam Tip: When \( \log(x \cos x) \) appears, remember to expand it as \( \log x + \log \cos x \) before differentiating to simplify the process. Also, be careful with the derivative of \( \frac{1}{x} \).

Question 12. \( x^y + y^x = 1 \)
Answer: Let \( u = x^y \) and \( v = y^x \).
The given equation is \( u + v = 1 \).
Differentiating implicitly with respect to \( x \), we get \( \frac{du}{dx} + \frac{dv}{dx} = 0 \).

First, find \( \frac{du}{dx} \) for \( u = x^y \).
Taking the natural logarithm of both sides: \( \log u = \log (x^y) \)
Using \( \log a^b = b \log a \): \( \log u = y \log x \)
Differentiating both sides with respect to \( x \) using the product rule, remembering \( y \) is a function of \( x \):
\( \frac{1}{u} \frac{du}{dx} = \frac{dy}{dx} \cdot \log x + y \cdot \frac{1}{x} \)
Multiply by \( u \): \( \frac{du}{dx} = u \left( \frac{dy}{dx} \log x + \frac{y}{x} \right) \)
Substitute back \( u = x^y \): \( \frac{du}{dx} = x^y \left( \frac{dy}{dx} \log x + \frac{y}{x} \right) \).

Next, find \( \frac{dv}{dx} \) for \( v = y^x \).
Taking the natural logarithm of both sides: \( \log v = \log (y^x) \)
Using \( \log a^b = b \log a \): \( \log v = x \log y \)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{v} \frac{dv}{dx} = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} \)
Multiply by \( v \): \( \frac{dv}{dx} = v \left( \log y + \frac{x}{y} \frac{dy}{dx} \right) \)
Substitute back \( v = y^x \): \( \frac{dv}{dx} = y^x \left( \log y + \frac{x}{y} \frac{dy}{dx} \right) \).

Now, substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into \( \frac{du}{dx} + \frac{dv}{dx} = 0 \):
\( x^y \left( \frac{dy}{dx} \log x + \frac{y}{x} \right) + y^x \left( \log y + \frac{x}{y} \frac{dy}{dx} \right) = 0 \)
\( x^y \frac{dy}{dx} \log x + x^y \frac{y}{x} + y^x \log y + y^x \frac{x}{y} \frac{dy}{dx} = 0 \)
\( \left( x^y \log x + y^x \frac{x}{y} \right) \frac{dy}{dx} = - \left( x^y \frac{y}{x} + y^x \log y \right) \)
\( \left( x^y \log x + x y^{x-1} \right) \frac{dy}{dx} = - \left( y x^{y-1} + y^x \log y \right) \)
\( \frac{dy}{dx} = - \frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}} \)
In simple words: This problem involves implicit differentiation with two complex terms. We differentiate each term separately using logarithms and the product rule, remembering that \( y \) is a function of \( x \). Then, we gather all \( \frac{dy}{dx} \) terms to one side to solve for it.

Exam Tip: For implicit differentiation of terms like \( x^y \) and \( y^x \), remember to use logarithmic differentiation. When differentiating \( \log y \), always include the \( \frac{dy}{dx} \) term from the chain rule.

Question 13. \( y^x = x^y \)
Answer: The given equation is \( y^x = x^y \).
Since both the base and exponent are variables, we use logarithmic differentiation.
Taking the natural logarithm of both sides, we get:
\( \log (y^x) = \log (x^y) \)
Using the logarithm property \( \log a^b = b \log a \):
\( x \log y = y \log x \)
Now, differentiate both sides with respect to \( x \) using the product rule, remembering that \( y \) is a function of \( x \):
\( \frac{d}{dx}(x \log y) = \frac{d}{dx}(y \log x) \)
\( 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \cdot \log x + y \cdot \frac{1}{x} \)
\( \log y + \frac{x}{y} \frac{dy}{dx} = \log x \frac{dy}{dx} + \frac{y}{x} \)
Group terms with \( \frac{dy}{dx} \) on one side and other terms on the other side:
\( \frac{x}{y} \frac{dy}{dx} - \log x \frac{dy}{dx} = \frac{y}{x} - \log y \)
Factor out \( \frac{dy}{dx} \):
\( \left( \frac{x}{y} - \log x \right) \frac{dy}{dx} = \frac{y}{x} - \log y \)
Simplify the expressions in the parentheses:
\( \left( \frac{x - y \log x}{y} \right) \frac{dy}{dx} = \frac{y - x \log y}{x} \)
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y}{x} \cdot \frac{y - x \log y}{x - y \log x} \)
\( \frac{dy}{dx} = \frac{y(y - x \log y)}{x(x - y \log x)} \)
In simple words: This equation implicitly defines \( y \) as a function of \( x \). We take logarithms on both sides to simplify the powers, then differentiate implicitly using the product rule. Finally, we rearrange the terms to solve for \( \frac{dy}{dx} \).

Exam Tip: For implicit differentiation of \( f(y)^x = g(x)^y \), taking logarithms is a must. Be careful when applying the product rule and remember to multiply by \( \frac{dy}{dx} \) for any derivative of \( y \).

Question 14. \( (\cos x)^y = (\cos y)^x \)
Answer: The given equation is \( (\cos x)^y = (\cos y)^x \).
Since both the base and exponent are variables, we use logarithmic differentiation.
Taking the natural logarithm of both sides, we get:
\( \log ((\cos x)^y) = \log ((\cos y)^x) \)
Using the logarithm property \( \log a^b = b \log a \):
\( y \log (\cos x) = x \log (\cos y) \)
Now, differentiate both sides with respect to \( x \) using the product rule, remembering that \( y \) is a function of \( x \):
\( \frac{d}{dx}(y \log (\cos x)) = \frac{d}{dx}(x \log (\cos y)) \)
\( \frac{dy}{dx} \cdot \log (\cos x) + y \cdot \frac{d}{dx}(\log (\cos x)) = 1 \cdot \log (\cos y) + x \cdot \frac{d}{dx}(\log (\cos y)) \)
\( \frac{dy}{dx} \log (\cos x) + y \cdot \frac{1}{\cos x} (-\sin x) = \log (\cos y) + x \cdot \frac{1}{\cos y} (-\sin y) \frac{dy}{dx} \)
\( \frac{dy}{dx} \log (\cos x) - y \tan x = \log (\cos y) - x \tan y \frac{dy}{dx} \)
Group terms with \( \frac{dy}{dx} \) on one side and other terms on the other side:
\( \frac{dy}{dx} \log (\cos x) + x \tan y \frac{dy}{dx} = \log (\cos y) + y \tan x \)
Factor out \( \frac{dy}{dx} \):
\( (\log (\cos x) + x \tan y) \frac{dy}{dx} = \log (\cos y) + y \tan x \)
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{\log (\cos y) + y \tan x}{\log (\cos x) + x \tan y} \)
In simple words: For this implicit equation, we take the natural logarithm of both sides. This helps to simplify the exponents. Then, we differentiate each side using the product rule and chain rule, collecting all terms with \( \frac{dy}{dx} \) to solve for it.

Exam Tip: Be careful with the chain rule for \( \log (\cos x) \) and \( \log (\cos y) \). The derivative of \( \log (\cos x) \) is \( \frac{-\sin x}{\cos x} = -\tan x \), and for \( \log (\cos y) \) it is \( \frac{-\sin y}{\cos y} \frac{dy}{dx} = -\tan y \frac{dy}{dx} \).

Question 15. \( xy = e^{x-y} \)
Answer: The given equation is \( xy = e^{x-y} \).
To differentiate this implicitly, taking the natural logarithm of both sides can simplify the exponential term.
\( \log (xy) = \log (e^{x-y}) \)
Using logarithm properties, \( \log(AB) = \log A + \log B \) and \( \log(e^k) = k \):
\( \log x + \log y = x - y \)
Now, differentiate both sides implicitly with respect to \( x \):
\( \frac{d}{dx}(\log x) + \frac{d}{dx}(\log y) = \frac{d}{dx}(x) - \frac{d}{dx}(y) \)
\( \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx} \)
Group terms with \( \frac{dy}{dx} \) on one side and other terms on the other side:
\( \frac{1}{y} \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x} \)
Factor out \( \frac{dy}{dx} \):
\( \left( \frac{1}{y} + 1 \right) \frac{dy}{dx} = \frac{x - 1}{x} \)
Simplify the expression in the parentheses:
\( \left( \frac{1 + y}{y} \right) \frac{dy}{dx} = \frac{x - 1}{x} \)
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y}{1 + y} \cdot \frac{x - 1}{x} \)
\( \frac{dy}{dx} = \frac{y(x - 1)}{x(y + 1)} \)
In simple words: We take the natural logarithm of both sides to get rid of the exponent. Then, we differentiate each term with respect to \( x \), remembering that \( y \) is a function of \( x \). Finally, we rearrange the equation to find \( \frac{dy}{dx} \).

Exam Tip: Always remember that \( \log e = 1 \). When differentiating \( \log y \), the chain rule gives \( \frac{1}{y} \frac{dy}{dx} \). Be careful with algebraic manipulation to isolate \( \frac{dy}{dx} \).

Question 16. Find the derivatives of the functions given by \( f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \) and hence find \( f' (1) \).
Answer: Let \( y = f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \).
Since this is a product of multiple functions, logarithmic differentiation is efficient.
Taking the natural logarithm of both sides, we get:
\( \log y = \log ((1 + x) (1 + x^2) (1 + x^4) (1 + x^8)) \)
Using the logarithm property \( \log(ABCD) = \log A + \log B + \log C + \log D \):
\( \log y = \log (1 + x) + \log (1 + x^2) + \log (1 + x^4) + \log (1 + x^8) \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log (1 + x)) + \frac{d}{dx}(\log (1 + x^2)) + \frac{d}{dx}(\log (1 + x^4)) + \frac{d}{dx}(\log (1 + x^8)) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{1 + x} \cdot 1 + \frac{1}{1 + x^2} \cdot 2x + \frac{1}{1 + x^4} \cdot 4x^3 + \frac{1}{1 + x^8} \cdot 8x^7 \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \)
Multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left( \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \right) \)
Substitute back \( y = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \):
\( f'(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \left( \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \right) \)

Now, we need to find \( f'(1) \). Substitute \( x = 1 \) into the expression for \( f'(x) \):
First, calculate \( y \) at \( x = 1 \):
\( y(1) = (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8) \)
\( y(1) = (2) (2) (2) (2) = 16 \).
Next, calculate the sum of fractions at \( x = 1 \):
\( \frac{1}{1 + 1} + \frac{2(1)}{1 + 1^2} + \frac{4(1)^3}{1 + 1^4} + \frac{8(1)^7}{1 + 1^8} \)
\( = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} \)
\( = \frac{1 + 2 + 4 + 8}{2} = \frac{15}{2} \).
Finally, multiply these two results:
\( f'(1) = 16 \cdot \frac{15}{2} = 8 \cdot 15 = 120 \).
In simple words: We used logarithms to make differentiating the product easier. First, we found the general derivative \( f'(x) \). Then, we plugged in \( x=1 \) into this derivative to find the specific value of \( f'(1) \).

Exam Tip: Logarithmic differentiation is very helpful for products of many terms. When calculating \( f'(a) \), it's often easiest to calculate \( f(a) \) and the sum of derivatives separately, then multiply them.

Question 17. Differentiate \( (x^2 – 5x + 8)(x^3 + 7x + 9) \) in three ways mentioned below?
(i) by using product rule.
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?
Answer: Let \( y = (x^2 – 5x + 8)(x^3 + 7x + 9) \).

**Method (i): Using the product rule**
The product rule states that if \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \).
Let \( u = x^2 – 5x + 8 \) and \( v = x^3 + 7x + 9 \).
Then \( u' = \frac{d}{dx}(x^2 – 5x + 8) = 2x – 5 \).
And \( v' = \frac{d}{dx}(x^3 + 7x + 9) = 3x^2 + 7 \).
Applying the product rule:
\( \frac{dy}{dx} = (2x – 5)(x^3 + 7x + 9) + (x^2 – 5x + 8)(3x^2 + 7) \)
Expand and simplify:
\( = (2x^4 + 14x^2 + 18x - 5x^3 - 35x - 45) + (3x^4 + 7x^2 - 15x^3 - 35x + 24x^2 + 56) \)
\( = 2x^4 - 5x^3 + 14x^2 - 17x - 45 + 3x^4 - 15x^3 + 31x^2 - 35x + 56 \)
Combine like terms:
\( = (2x^4 + 3x^4) + (-5x^3 - 15x^3) + (14x^2 + 31x^2) + (-17x - 35x) + (-45 + 56) \)
\( = 5x^4 - 20x^3 + 45x^2 - 52x + 11 \).

**Method (ii): Expanding the product to obtain a single polynomial**
First, expand the product \( y = (x^2 – 5x + 8)(x^3 + 7x + 9) \):
\( y = x^2(x^3 + 7x + 9) - 5x(x^3 + 7x + 9) + 8(x^3 + 7x + 9) \)
\( y = (x^5 + 7x^3 + 9x^2) - (5x^4 + 35x^2 + 45x) + (8x^3 + 56x + 72) \)
\( y = x^5 - 5x^4 + (7x^3 + 8x^3) + (9x^2 - 35x^2) + (-45x + 56x) + 72 \)
\( y = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72 \).
Now, differentiate the polynomial term by term:
\( \frac{dy}{dx} = \frac{d}{dx}(x^5) - \frac{d}{dx}(5x^4) + \frac{d}{dx}(15x^3) - \frac{d}{dx}(26x^2) + \frac{d}{dx}(11x) + \frac{d}{dx}(72) \)
\( \frac{dy}{dx} = 5x^4 - 5(4x^3) + 15(3x^2) - 26(2x) + 11(1) + 0 \)
\( \frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11 \).

**Method (iii): Using logarithmic differentiation**
Let \( y = (x^2 – 5x + 8)(x^3 + 7x + 9) \).
Taking the natural logarithm of both sides:
\( \log y = \log ((x^2 – 5x + 8)(x^3 + 7x + 9)) \)
Using the logarithm property \( \log(AB) = \log A + \log B \):
\( \log y = \log (x^2 – 5x + 8) + \log (x^3 + 7x + 9) \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log (x^2 – 5x + 8)) + \frac{d}{dx}(\log (x^3 + 7x + 9)) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x^2 – 5x + 8} \cdot (2x – 5) + \frac{1}{x^3 + 7x + 9} \cdot (3x^2 + 7) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{2x – 5}{x^2 – 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \)
Multiply both sides by \( y \):
\( \frac{dy}{dx} = y \left( \frac{2x – 5}{x^2 – 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right) \)
Substitute back \( y = (x^2 – 5x + 8)(x^3 + 7x + 9) \):
\( \frac{dy}{dx} = (x^2 – 5x + 8)(x^3 + 7x + 9) \left( \frac{2x – 5}{x^2 – 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right) \)
Distribute \( y \) into the parentheses:
\( \frac{dy}{dx} = (x^3 + 7x + 9)(2x – 5) + (x^2 – 5x + 8)(3x^2 + 7) \)
This is the same expression obtained from Method (i). Expanding and simplifying this will yield the same result:
\( \frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11 \).

**Conclusion:** All three methods give the same answer: \( 5x^4 - 20x^3 + 45x^2 - 52x + 11 \).
In simple words: We used three different ways to find the derivative: the product rule, expanding the expression first, and using logarithms. All methods produced the exact same final answer, showing that different approaches can lead to the same correct result.

Exam Tip: This question demonstrates that different differentiation techniques should yield the same result for a given function. Choose the method that you find most straightforward and least prone to errors for complex expressions.

Question 18. If u, v and w are functions of x, then show that \( \frac{d}{dx}(u \cdot v \cdot w) = \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx} \) in two ways : first by repeated application of product rule and secondly by logarithmic differentiation.
Answer: Let \( y = u \cdot v \cdot w \), where \( u, v, w \) are functions of \( x \).

**Method (i): Repeated application of the product rule**
We can group the terms as \( y = u \cdot (vw) \).
Applying the product rule \( (fg)' = f'g + fg' \) where \( f=u \) and \( g=vw \):
\( \frac{dy}{dx} = \frac{du}{dx} \cdot (vw) + u \cdot \frac{d}{dx}(vw) \)
Now, apply the product rule again for \( \frac{d}{dx}(vw) \):
\( \frac{d}{dx}(vw) = \frac{dv}{dx} \cdot w + v \cdot \frac{dw}{dx} \)
Substitute this back into the equation for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{du}{dx} \cdot vw + u \left( \frac{dv}{dx} \cdot w + v \cdot \frac{dw}{dx} \right) \)
\( \frac{dy}{dx} = \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx} \).
This proves the formula.

**Method (ii): Logarithmic differentiation**
Let \( y = u \cdot v \cdot w \).
Taking the natural logarithm of both sides:
\( \log y = \log (u \cdot v \cdot w) \)
Using the logarithm property \( \log(ABC) = \log A + \log B + \log C \):
\( \log y = \log u + \log v + \log w \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log u) + \frac{d}{dx}(\log v) + \frac{d}{dx}(\log w) \)
Applying the chain rule for each term (e.g., \( \frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx} \)):
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \)
Multiply both sides by \( y \):
\( \frac{dy}{dx} = y \left( \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \right) \)
Substitute back \( y = u \cdot v \cdot w \):
\( \frac{dy}{dx} = (u \cdot v \cdot w) \left( \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \right) \)
Distribute \( u \cdot v \cdot w \) into the parentheses:
\( \frac{dy}{dx} = \frac{u \cdot v \cdot w}{u} \frac{du}{dx} + \frac{u \cdot v \cdot w}{v} \frac{dv}{dx} + \frac{u \cdot v \cdot w}{w} \frac{dw}{dx} \)
\( \frac{dy}{dx} = v \cdot w \frac{du}{dx} + u \cdot w \frac{dv}{dx} + u \cdot v \frac{dw}{dx} \).
This also proves the formula.
In simple words: We showed the derivative formula for a product of three functions in two ways. First, by applying the product rule twice. Second, by using logarithms to change the product into a sum, differentiating, and then multiplying back. Both ways arrive at the same important formula.

Exam Tip: This formula is a generalization of the product rule for three functions. Logarithmic differentiation often provides a cleaner and more systematic way to derive such formulas for products of many terms.