Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.5

Question 1. Differentiate the following with respect to x.
(i) \( \frac{5}{x^{4}}-\frac{2}{x^{3}}+\frac{5}{x} \)
(ii) \( 3x^4 - 2x^3 + x + 8 \)
(iii) \( \sqrt{x}+\frac{1}{\sqrt[3]{x}}+e^{x} \)
(iv) \( \frac{3+2 x-x^{2}}{x} \)
(v) \( x^3 e^x \)
(vi) \( (x^2 - 3x + 2) (x + 1) \)
(vii) \( x^4 - 3 \sin x + \cos x \)
(viii) \( \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2} \)
Answer:
(i) Let \( y = \frac{5}{x^{4}}-\frac{2}{x^{3}}+\frac{5}{x} \)
We can rewrite this as: \( y = 5x^{-4} - 2x^{-3} + 5x^{-1} \)
Now, we differentiate with respect to x using the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \):
\( \frac{dy}{dx} = 5 \frac{d}{dx}(x^{-4}) - 2 \frac{d}{dx}(x^{-3}) + 5 \frac{d}{dx}(x^{-1}) \)
\( = 5(-4x^{-4-1}) - 2(-3x^{-3-1}) + 5(-1x^{-1-1}) \)
\( = -20x^{-5} + 6x^{-4} - 5x^{-2} \)
Finally, we write the answer with positive exponents:
\( = -\frac{20}{x^5} + \frac{6}{x^4} - \frac{5}{x^2} \)
(ii) Let \( y = 3x^4 - 2x^3 + x + 8 \)
Differentiate each term separately:
\( \frac{dy}{dx} = \frac{d}{dx}(3x^4) - \frac{d}{dx}(2x^3) + \frac{d}{dx}(x) + \frac{d}{dx}(8) \)
\( = 3(4x^{4-1}) - 2(3x^{3-1}) + 1x^{1-1} + 0 \)
\( = 12x^3 - 6x^2 + 1 \)
(iii) Let \( y = \sqrt{x}+\frac{1}{\sqrt[3]{x}}+e^{x} \)
Rewrite the terms with fractional exponents: \( y = x^{\frac{1}{2}} + x^{-\frac{1}{3}} + e^x \)
Now, differentiate each term using the power rule and the derivative of \( e^x \):
\( \frac{dy}{dx} = \frac{d}{dx}(x^{\frac{1}{2}}) + \frac{d}{dx}(x^{-\frac{1}{3}}) + \frac{d}{dx}(e^x) \)
\( = \frac{1}{2}x^{\frac{1}{2}-1} + (-\frac{1}{3})x^{-\frac{1}{3}-1} + e^x \)
\( = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{3}x^{-\frac{4}{3}} + e^x \)
Write the answer with positive exponents and radical form:
\( = \frac{1}{2\sqrt{x}} - \frac{1}{3x^{\frac{4}{3}}} + e^x \)
(iv) Let \( y = \frac{3+2 x-x^{2}}{x} \)
First, simplify the expression by dividing each term by x:
\( y = \frac{3}{x} + \frac{2x}{x} - \frac{x^2}{x} \)
\( y = 3x^{-1} + 2 - x \)
Now, differentiate each term with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(3x^{-1}) + \frac{d}{dx}(2) - \frac{d}{dx}(x) \)
\( = 3(-1x^{-1-1}) + 0 - 1 \)
\( = -3x^{-2} - 1 \)
Write the answer with positive exponents:
\( = -\frac{3}{x^2} - 1 \)
(v) Let \( y = x^3 e^x \)
Here, we use the product rule, which states that if \( y = uv \), then \( \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \).
Let \( u = x^3 \) and \( v = e^x \).
Then \( \frac{du}{dx} = 3x^2 \) and \( \frac{dv}{dx} = e^x \).
\( \frac{dy}{dx} = x^3(e^x) + e^x(3x^2) \)
\( = x^3e^x + 3x^2e^x \)
Factor out \( e^x \) or \( x^2e^x \):
\( = e^x(x^3 + 3x^2) \)
\( = x^2e^x(x+3) \)
(vi) Let \( y = (x^2 - 3x + 2)(x + 1) \)
First, expand the expression by multiplying the two factors:
\( y = x^2(x) + x^2(1) - 3x(x) - 3x(1) + 2(x) + 2(1) \)
\( y = x^3 + x^2 - 3x^2 - 3x + 2x + 2 \)
Combine like terms:
\( y = x^3 - 2x^2 - x + 2 \)
Now, differentiate each term with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x^2) - \frac{d}{dx}(x) + \frac{d}{dx}(2) \)
\( = 3x^2 - 2(2x) - 1 + 0 \)
\( = 3x^2 - 4x - 1 \)
(vii) Let \( y = x^4 - 3 \sin x + \cos x \)
Differentiate each term separately:
\( \frac{dy}{dx} = \frac{d}{dx}(x^4) - 3\frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x) \)
\( = 4x^3 - 3(\cos x) + (-\sin x) \)
\( = 4x^3 - 3 \cos x - \sin x \)
(viii) Let \( y = \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2} \)
First, expand the square: \( (a+b)^2 = a^2 + 2ab + b^2 \).
Here, \( a = \sqrt{x} \) and \( b = \frac{1}{\sqrt{x}} \).
\( y = (\sqrt{x})^2 + 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) + \left(\frac{1}{\sqrt{x}}\right)^2 \)
\( y = x + 2(1) + \frac{1}{x} \)
\( y = x + 2 + x^{-1} \)
Now, differentiate each term with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(2) + \frac{d}{dx}(x^{-1}) \)
\( = 1 + 0 + (-1x^{-1-1}) \)
\( = 1 - x^{-2} \)
Write the answer with positive exponents:
\( = 1 - \frac{1}{x^2} \)
In simple words: For each part, we find the rate at which the function changes. We use basic rules like the power rule, product rule, and quotient rule, along with derivatives of common functions like \( e^x \), \( \sin x \), and \( \cos x \). Sometimes, simplifying the function first makes differentiation easier.

🎯 Exam Tip: Always simplify the expression as much as possible before differentiating, especially for fractions or powers. Remember to apply the correct differentiation rule for each type of function (sum/difference, product, quotient, chain rule).

Question 2. Differentiate the following with respect to x.
(i) \( \frac{e^{x}}{1+x} \)
(ii) \( \frac{x^{2}+x+1}{x^{2}-x+1} \)
(iii) \( \frac{e^{x}}{1+e^{x}} \)
Answer:
(i) Let \( y = \frac{e^{x}}{1+x} \)
We use the quotient rule: If \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \).
Let \( u = e^x \) and \( v = 1+x \).
Then \( \frac{du}{dx} = e^x \) and \( \frac{dv}{dx} = 1 \).
Substitute these into the quotient rule formula:
\( \frac{dy}{dx} = \frac{(1+x)\frac{d}{dx}(e^x) - e^x\frac{d}{dx}(1+x)}{(1+x)^2} \)
\( = \frac{(1+x)e^x - e^x(1)}{(1+x)^2} \)
Factor out \( e^x \) from the numerator:
\( = \frac{e^x((1+x)-1)}{(1+x)^2} \)
\( = \frac{e^x(x)}{(1+x)^2} \)
\( = \frac{xe^x}{(1+x)^2} \)
(ii) Let \( y = \frac{x^{2}+x+1}{x^{2}-x+1} \)
Again, we use the quotient rule. Let \( u = x^2+x+1 \) and \( v = x^2-x+1 \).
Then \( \frac{du}{dx} = 2x+1 \) and \( \frac{dv}{dx} = 2x-1 \).
Substitute these into the quotient rule formula:
\( \frac{dy}{dx} = \frac{(x^2-x+1)\frac{d}{dx}(x^2+x+1) - (x^2+x+1)\frac{d}{dx}(x^2-x+1)}{(x^2-x+1)^2} \)
\( = \frac{(x^2-x+1)(2x+1) - (x^2+x+1)(2x-1)}{(x^2-x+1)^2} \)
Expand the terms in the numerator:
\( = \frac{(2x^3+x^2-2x^2-x+2x+1) - (2x^3-x^2+2x^2-x+2x-1)}{(x^2-x+1)^2} \)
\( = \frac{(2x^3-x^2+x+1) - (2x^3+x^2+x-1)}{(x^2-x+1)^2} \)
Now, subtract the second expanded term from the first:
\( = \frac{2x^3-x^2+x+1 - 2x^3-x^2-x+1}{(x^2-x+1)^2} \)
Combine like terms in the numerator:
\( = \frac{-2x^2+2}{(x^2-x+1)^2} \)
Factor out -2 from the numerator:
\( = \frac{-2(x^2-1)}{(x^2-x+1)^2} \)
This can also be written as:
\( = \frac{2(1-x^2)}{(x^2-x+1)^2} \)
(iii) Let \( y = \frac{e^{x}}{1+e^{x}} \)
Using the quotient rule: Let \( u = e^x \) and \( v = 1+e^x \).
Then \( \frac{du}{dx} = e^x \) and \( \frac{dv}{dx} = e^x \).
Substitute these into the quotient rule formula:
\( \frac{dy}{dx} = \frac{(1+e^x)\frac{d}{dx}(e^x) - e^x\frac{d}{dx}(1+e^x)}{(1+e^x)^2} \)
\( = \frac{(1+e^x)e^x - e^x(e^x)}{(1+e^x)^2} \)
Factor out \( e^x \) from the numerator:
\( = \frac{e^x((1+e^x)-e^x)}{(1+e^x)^2} \)
\( = \frac{e^x(1)}{(1+e^x)^2} \)
\( = \frac{e^x}{(1+e^x)^2} \)
In simple words: When a function is a fraction (one function divided by another), we use a special rule called the quotient rule to find its derivative. This rule helps us differentiate these types of complex expressions step-by-step.

🎯 Exam Tip: The quotient rule can be tricky with negative signs and complex algebraic simplification. Always double-check your calculations, especially when distributing negative signs after expansion.

Question 3. Differentiate the following with respect to x.
(i) \( x \sin x \)
(ii) \( e^x \sin x \)
(iii) \( e^x (x + \log x) \)
(iv) \( \sin x \cos x \)
(v) \( x^3 e^x \)
Answer:
(i) Let \( y = x \sin x \)
We use the product rule: If \( y = uv \), then \( \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \).
Let \( u = x \) and \( v = \sin x \).
Then \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \cos x \).
Substitute these into the product rule formula:
\( \frac{dy}{dx} = x(\cos x) + \sin x(1) \)
\( = x \cos x + \sin x \)
(ii) Let \( y = e^x \sin x \)
Using the product rule. Let \( u = e^x \) and \( v = \sin x \).
Then \( \frac{du}{dx} = e^x \) and \( \frac{dv}{dx} = \cos x \).
Substitute these into the product rule formula:
\( \frac{dy}{dx} = e^x(\cos x) + \sin x(e^x) \)
\( = e^x \cos x + e^x \sin x \)
Factor out \( e^x \):
\( = e^x(\cos x + \sin x) \)
(iii) Let \( y = e^x (x + \log x) \)
Using the product rule. Let \( u = e^x \) and \( v = x + \log x \).
Then \( \frac{du}{dx} = e^x \) and \( \frac{dv}{dx} = 1 + \frac{1}{x} \).
Substitute these into the product rule formula:
\( \frac{dy}{dx} = e^x\left(1 + \frac{1}{x}\right) + (x + \log x)e^x \)
Factor out \( e^x \):
\( = e^x \left(1 + \frac{1}{x} + x + \log x\right) \)
(iv) Let \( y = \sin x \cos x \)
Method 1: Using the product rule.
Let \( u = \sin x \) and \( v = \cos x \).
Then \( \frac{du}{dx} = \cos x \) and \( \frac{dv}{dx} = -\sin x \).
Substitute these into the product rule formula:
\( \frac{dy}{dx} = \sin x(-\sin x) + \cos x(\cos x) \)
\( = -\sin^2 x + \cos^2 x \)
This can be written using a trigonometric identity as:
\( = \cos^2 x - \sin^2 x = \cos(2x) \)
Method 2: Using a trigonometric identity first.
We know that \( \sin(2x) = 2 \sin x \cos x \). So, \( \sin x \cos x = \frac{1}{2} \sin(2x) \).
Let \( y = \frac{1}{2} \sin(2x) \).
Now differentiate using the chain rule:
\( \frac{dy}{dx} = \frac{1}{2} \cos(2x) \cdot \frac{d}{dx}(2x) \)
\( = \frac{1}{2} \cos(2x) \cdot 2 \)
\( = \cos(2x) \)
Both methods lead to the same result, showing how trigonometric identities can simplify calculations.
(v) Let \( y = x^3 e^x \)
Using the product rule. Let \( u = x^3 \) and \( v = e^x \).
Then \( \frac{du}{dx} = 3x^2 \) and \( \frac{dv}{dx} = e^x \).
Substitute these into the product rule formula:
\( \frac{dy}{dx} = x^3(e^x) + e^x(3x^2) \)
\( = x^3e^x + 3x^2e^x \)
Factor out \( x^2e^x \):
\( = x^2e^x(x+3) \)
In simple words: When two functions are multiplied together, we use the product rule to find their derivative. It helps us differentiate expressions that have a multiplication in them. Sometimes, simplifying with identities first can make the process easier.

🎯 Exam Tip: For product rule problems, correctly identifying \( u \) and \( v \) and their derivatives is crucial. Remember to fully expand and simplify the expression after applying the rule.

Question 4. Differentiate the following with respect to x.
(i) \( \sin^2 x \)
(ii) \( \cos^2 x \)
(iii) \( \cos^3 x \)
(iv) \( \sqrt{1+x^{2}} \)
(v) \( (ax^2 + bx + c)^n \)
(vi) \( \sin(x^2) \)
(vii) \( \frac{1}{\sqrt{1+x^{2}}} \)
Answer:
For the following problems, the chain rule will be used: \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \). For power functions, this means \( \frac{d}{dx}[f(x)]^n = n[f(x)]^{n-1} \cdot \frac{d}{dx}f(x) \).
(i) Let \( y = \sin^2 x \)
We can write this as \( y = (\sin x)^2 \).
Here, \( f(x) = \sin x \) and \( n = 2 \).
Using the chain rule:
\( \frac{dy}{dx} = 2(\sin x)^{2-1} \cdot \frac{d}{dx}(\sin x) \)
\( = 2 \sin x (\cos x) \)
Using the identity \( \sin(2x) = 2 \sin x \cos x \), we can write:
\( = \sin(2x) \)
(ii) Let \( y = \cos^2 x \)
We can write this as \( y = (\cos x)^2 \).
Here, \( f(x) = \cos x \) and \( n = 2 \).
Using the chain rule:
\( \frac{dy}{dx} = 2(\cos x)^{2-1} \cdot \frac{d}{dx}(\cos x) \)
\( = 2 \cos x (-\sin x) \)
\( = -2 \sin x \cos x \)
Using the identity \( -\sin(2x) = -2 \sin x \cos x \), we can write:
\( = -\sin(2x) \)
(iii) Let \( y = \cos^3 x \)
We can write this as \( y = (\cos x)^3 \).
Here, \( f(x) = \cos x \) and \( n = 3 \).
Using the chain rule:
\( \frac{dy}{dx} = 3(\cos x)^{3-1} \cdot \frac{d}{dx}(\cos x) \)
\( = 3 \cos^2 x (-\sin x) \)
\( = -3 \cos^2 x \sin x \)
This can also be expressed using a double angle identity. Multiply and divide by 2:
\( = -\frac{3}{2} \cos x (2 \sin x \cos x) \)
\( = -\frac{3}{2} \cos x \sin(2x) \)
(iv) Let \( y = \sqrt{1+x^{2}} \)
We can write this as \( y = (1+x^2)^{\frac{1}{2}} \).
Here, the outer function is \( f(u) = u^{\frac{1}{2}} \) and the inner function is \( g(x) = 1+x^2 \).
First, find the derivative of the outer function with respect to \( u \): \( f'(u) = \frac{1}{2}u^{-\frac{1}{2}} \).
Next, find the derivative of the inner function with respect to \( x \): \( g'(x) = \frac{d}{dx}(1+x^2) = 0 + 2x = 2x \).
Apply the chain rule \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \):
\( \frac{dy}{dx} = \frac{1}{2}(1+x^2)^{-\frac{1}{2}} \cdot (2x) \)
\( = \frac{1}{2\sqrt{1+x^2}} \cdot (2x) \)
\( = \frac{x}{\sqrt{1+x^2}} \)
(v) Let \( y = (ax^2 + bx + c)^n \)
Here, the outer function is \( f(u) = u^n \) and the inner function is \( g(x) = ax^2 + bx + c \).
First, find the derivative of the outer function with respect to \( u \): \( f'(u) = nu^{n-1} \).
Next, find the derivative of the inner function with respect to \( x \): \( g'(x) = \frac{d}{dx}(ax^2+bx+c) = 2ax + b \).
Apply the chain rule:
\( \frac{dy}{dx} = n(ax^2+bx+c)^{n-1} \cdot (2ax+b) \)
(vi) Let \( y = \sin(x^2) \)
Here, the outer function is \( f(u) = \sin u \) and the inner function is \( g(x) = x^2 \).
First, find the derivative of the outer function with respect to \( u \): \( f'(u) = \cos u \).
Next, find the derivative of the inner function with respect to \( x \): \( g'(x) = \frac{d}{dx}(x^2) = 2x \).
Apply the chain rule:
\( \frac{dy}{dx} = \cos(x^2) \cdot (2x) \)
\( = 2x \cos(x^2) \)
(vii) Let \( y = \frac{1}{\sqrt{1+x^{2}}} \)
We can write this as \( y = (1+x^2)^{-\frac{1}{2}} \).
Here, the outer function is \( f(u) = u^{-\frac{1}{2}} \) and the inner function is \( g(x) = 1+x^2 \).
First, find the derivative of the outer function with respect to \( u \): \( f'(u) = -\frac{1}{2}u^{-\frac{3}{2}} \).
Next, find the derivative of the inner function with respect to \( x \): \( g'(x) = \frac{d}{dx}(1+x^2) = 2x \).
Apply the chain rule:
\( \frac{dy}{dx} = -\frac{1}{2}(1+x^2)^{-\frac{3}{2}} \cdot (2x) \)
\( = -x(1+x^2)^{-\frac{3}{2}} \)
Write the answer with positive exponents and radical form:
\( = \frac{-x}{(1+x^2)^{\frac{3}{2}}} \)
\( = \frac{-x}{(1+x^2)\sqrt{1+x^2}} \)
In simple words: The chain rule is used when one function is "nested" inside another, like \( (x^2+1)^5 \) or \( \sin(2x) \). It helps us differentiate these composite functions by taking the derivative of the outer function first, then multiplying it by the derivative of the inner function.

🎯 Exam Tip: Remember to clearly identify the "outer" and "inner" functions when applying the chain rule. A common mistake is to forget to multiply by the derivative of the inner function, which is often crucial for the correct answer.