Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.7

Question 1. Differentiate the following with respect to x.
(i) \( y = x^{\sin x} \)
(ii) \( y = (\sin x)^x \)
(iii) \( y = (\sin x)^{\tan x} \)
(iv) \( y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x^{2}+x+1)}} \)
Answer:
(i) Let the given function be \( y = x^{\sin x} \).
To differentiate this, we take the natural logarithm on both sides.
\( \log y = \log(x^{\sin x}) \)
Using logarithm properties, the exponent comes down.
\( \log y = \sin x \log x \)
Now, differentiate both sides with respect to \( x \). We use the product rule on the right side.
\( \frac{1}{y} \frac{dy}{dx} = \sin x \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(\sin x) \)
\( \frac{1}{y} \frac{dy}{dx} = \sin x \left(\frac{1}{x}\right) + \log x (\cos x) \)
To find \( \frac{dy}{dx} \), multiply both sides by \( y \).
\( \frac{dy}{dx} = y\left[\frac{\sin x}{x} + \cos x \log x\right] \)
Finally, substitute back the value of \( y \).
\( \frac{dy}{dx} = x^{\sin x}\left[\frac{\sin x}{x} + \cos x \log x\right] \)

(ii) Let the given function be \( y = (\sin x)^x \).
Take the natural logarithm on both sides to simplify.
\( \log y = \log((\sin x)^x) \)
Using logarithm rules, the power \( x \) moves to the front.
\( \log y = x \log(\sin x) \)
Differentiate both sides with respect to \( x \), applying the product rule on the right side.
\( \frac{1}{y} \frac{dy}{dx} = x \frac{d}{dx}(\log(\sin x)) + \log(\sin x) \frac{d}{dx}(x) \)
The derivative of \( \log(\sin x) \) is \( \frac{1}{\sin x} \cdot \cos x \). The derivative of \( x \) is 1.
\( \frac{1}{y} \frac{dy}{dx} = x \left(\frac{1}{\sin x} \cos x\right) + \log(\sin x) (1) \)
\( \frac{1}{y} \frac{dy}{dx} = x \cot x + \log(\sin x) \)
Multiply both sides by \( y \) to solve for \( \frac{dy}{dx} \).
\( \frac{dy}{dx} = y[x \cot x + \log(\sin x)] \)
Substitute \( y = (\sin x)^x \) back into the equation.
\( \frac{dy}{dx} = (\sin x)^x[x \cot x + \log(\sin x)] \)

(iii) Let the function be \( y = (\sin x)^{\tan x} \).
Take the natural logarithm on both sides.
\( \log y = \log((\sin x)^{\tan x}) \)
Applying the power rule for logarithms, the exponent \( \tan x \) comes down.
\( \log y = \tan x \log(\sin x) \)
Now, differentiate both sides with respect to \( x \), using the product rule on the right side.
\( \frac{1}{y} \frac{dy}{dx} = \tan x \frac{d}{dx}(\log(\sin x)) + \log(\sin x) \frac{d}{dx}(\tan x) \)
The derivative of \( \log(\sin x) \) is \( \frac{1}{\sin x} \cos x = \cot x \). The derivative of \( \tan x \) is \( \sec^2 x \).
\( \frac{1}{y} \frac{dy}{dx} = \tan x \left(\frac{1}{\sin x} \cos x\right) + \log(\sin x) \sec^2 x \)
Simplify \( \tan x \left(\frac{\cos x}{\sin x}\right) = \frac{\sin x}{\cos x} \cdot \frac{\cos x}{\sin x} = 1 \).
\( \frac{1}{y} \frac{dy}{dx} = 1 + \log(\sin x) \sec^2 x \)
Multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \).
\( \frac{dy}{dx} = y[1 + \sec^2 x \log(\sin x)] \)
Substitute \( y = (\sin x)^{\tan x} \) back into the expression.
\( \frac{dy}{dx} = (\sin x)^{\tan x}[1 + \sec^2 x \log(\sin x)] \)

(iv) Let the given function be \( y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x^{2}+x+1)}} \).
First, rewrite the square root as a power of 1/2.
\( y = \left(\frac{(x-1)(x-2)}{(x-3)(x^{2}+x+1)}\right)^{1/2} \)
Take the natural logarithm on both sides to simplify the expression for differentiation.
\( \log y = \log \left(\left(\frac{(x-1)(x-2)}{(x-3)(x^{2}+x+1)}\right)^{1/2}\right) \)
Using logarithm properties, the power \( \frac{1}{2} \) moves to the front, and division becomes subtraction of logs, while multiplication becomes addition.
\( \log y = \frac{1}{2} \left[\log((x-1)(x-2)) - \log((x-3)(x^{2}+x+1))\right] \)
\( \log y = \frac{1}{2} \left[\log(x-1) + \log(x-2) - (\log(x-3) + \log(x^{2}+x+1))\right] \)
\( \log y = \frac{1}{2} \left[\log(x-1) + \log(x-2) - \log(x-3) - \log(x^{2}+x+1)\right] \)
Now, differentiate both sides with respect to \( x \). The derivative of \( \log u \) is \( \frac{1}{u} \frac{du}{dx} \).
For \( \log(x^{2}+x+1) \), its derivative is \( \frac{1}{x^{2}+x+1} (2x+1) \).
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[\frac{1}{x-1} \cdot 1 + \frac{1}{x-2} \cdot 1 - \frac{1}{x-3} \cdot 1 - \frac{1}{x^{2}+x+1} (2x+1)\right] \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[\frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{2x+1}{x^{2}+x+1}\right] \)
To find \( \frac{dy}{dx} \), multiply both sides by \( y \).
\( \frac{dy}{dx} = \frac{y}{2} \left[\frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{2x+1}{x^{2}+x+1}\right] \)
Substitute back the original expression for \( y \).
\( \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x^{2}+x+1)}} \left[\frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{2x+1}{x^{2}+x+1}\right] \)
In simple words: For questions like these, where a function is raised to another function, we use a trick called logarithmic differentiation. We take the log on both sides to bring the power down, then differentiate and solve for dy/dx. For complex fractions under a square root, logs help break them into simpler additions and subtractions before differentiating. This method makes difficult derivatives much easier to calculate.

🎯 Exam Tip: Remember to use logarithmic differentiation when variables are in both the base and the exponent, or for very complex products and quotients. Always substitute back the value of y at the end.

Question 2. If \( x^m \cdot y^n = (x+y)^{m+n} \), then show that \( \frac{dy}{dx} = \frac{y}{x} \).
Answer:
We are given the equation: \( x^m y^n = (x+y)^{m+n} \).
To simplify this expression and make differentiation easier, we take the natural logarithm on both sides.
\( \log(x^m y^n) = \log((x+y)^{m+n}) \)
Using logarithm properties, multiplication becomes addition of logs, and exponents come down as multipliers.
\( \log(x^m) + \log(y^n) = (m+n) \log(x+y) \)
\( m \log x + n \log y = (m+n) \log(x+y) \)
Now, differentiate both sides with respect to \( x \). Remember to use the chain rule for \( \log y \) and \( \log(x+y) \).
\( m \frac{d}{dx}(\log x) + n \frac{d}{dx}(\log y) = (m+n) \frac{d}{dx}(\log(x+y)) \)
\( m \left(\frac{1}{x}\right) + n \left(\frac{1}{y} \frac{dy}{dx}\right) = (m+n) \left(\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right)\right) \)
This gives:
\( \frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} + \frac{m+n}{x+y} \frac{dy}{dx} \)
Next, group the terms containing \( \frac{dy}{dx} \) on one side and other terms on the other side.
\( \frac{n}{y} \frac{dy}{dx} - \frac{m+n}{x+y} \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x} \)
Factor out \( \frac{dy}{dx} \) on the left side and find a common denominator on both sides.
\( \frac{dy}{dx} \left(\frac{n}{y} - \frac{m+n}{x+y}\right) = \frac{m+n}{x+y} - \frac{m}{x} \)
\( \frac{dy}{dx} \left(\frac{n(x+y) - y(m+n)}{y(x+y)}\right) = \frac{x(m+n) - m(x+y)}{x(x+y)} \)
Expand the terms in the numerators:
\( \frac{dy}{dx} \left(\frac{nx+ny-my-ny}{y(x+y)}\right) = \frac{mx+nx-mx-my}{x(x+y)} \)
Simplify the numerators:
\( \frac{dy}{dx} \left(\frac{nx-my}{y(x+y)}\right) = \frac{nx-my}{x(x+y)} \)
Now, solve for \( \frac{dy}{dx} \). We can cancel the common term \( (x+y) \) from the denominators and \( (nx-my) \) from the numerators (assuming \( nx-my \neq 0 \)).
\( \frac{dy}{dx} \left(\frac{nx-my}{y}\right) = \left(\frac{nx-my}{x}\right) \)
\( \frac{dy}{dx} = \frac{y}{x} \)
Thus, it is shown that \( \frac{dy}{dx} = \frac{y}{x} \). This type of problem is a classic application of implicit differentiation using logarithms.
In simple words: When we have a complex equation like this, especially with powers, taking the logarithm on both sides helps a lot. It turns multiplication into addition and brings down the powers, making it easier to differentiate. After doing that, we just collect all the terms with dy/dx and solve for it, which leads us to the answer y/x.

🎯 Exam Tip: For problems involving products of variables raised to powers, always use logarithmic differentiation. Pay close attention to algebraic simplification after differentiation to reach the final desired form.