Maharashtra Board Class 12 Maths Part 1 Chapter 3 Differentiation 3.3 Solutions

1. Find \( \frac{dy}{dx} \) if:

Question 1. \( y = x^{x^2} \)
Answer:
Solution:
\( y = x^{x^2} \)
\( \therefore \log y = \log x^{x^2} = x^{2x} \cdot \log x \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (x^{2x} \cdot \log x) \)

\( \implies x^{2x} \frac{d}{dx} (\log x) + (\log x) \frac{d}{dx} (x^{2x}) \)

\( \implies x^{2x} \cdot \frac{1}{x} + (\log x) \frac{d}{dx} (x^{2x}) \)

\( \implies \frac{dy}{dx} = y \left[ \frac{x^{2x}}{x} + (\log x) \frac{d}{dx} (x^{2x}) \right] \)

\( \implies \frac{dy}{dx} = x^{x^2} \left[ \frac{x^{2x}}{x} + (\log x) \frac{d}{dx} (x^{2x}) \right] \quad \ldots (1) \)
Let \( u = x^{2x} \)
Then \( \log u = \log x^{2x} = 2x \log x \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{u} \frac{du}{dx} = 2 \frac{d}{dx} (x \log x) \)

\( \implies 2 \left[ x \frac{d}{dx} (\log x) + (\log x) \frac{d}{dx} (x) \right] \)

\( \implies 2 \left[ x \cdot \frac{1}{x} + (\log x) \cdot 1 \right] \)

\( \implies 2 [1 + \log x] \)

\( \therefore \frac{du}{dx} = 2u(1 + \log x) \)

\( \therefore \frac{d}{dx} (x^{2x}) = 2x^{2x}(1 + \log x) \)

\( \therefore \text{ from (1),} \)
\( \frac{dy}{dx} = x^{x^2} \left[ \frac{x^{2x}}{x} + (\log x) \cdot 2x^{2x}(1 + \log x) \right] \)

\( \implies x^{x^2} \cdot x^{2x} \left[ \frac{1}{x} + (\log x) \cdot 2(1 + \log x) \right] \)
In simple words: To differentiate this function with a variable base and exponent, logarithmic differentiation is used. The function is broken down using properties of logarithms, and then chain rule is applied to its components, especially for the complex exponent part.

🎯 Exam Tip: Always clearly state when you are applying logarithmic properties and when you are differentiating. Pay close attention to the chain rule for nested functions.

Question 2. \( y = x^{e^x} \)
Answer:
Solution:
\( y = x^{e^x} \)
\( \therefore \log y = \log x^{e^x} = e^x \cdot \log x \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (e^x \cdot \log x) \)

\( \implies e^x \frac{d}{dx} (\log x) + (\log x) \frac{d}{dx} (e^x) \)

\( \implies e^x \cdot \frac{1}{x} + (\log x) (e^x) \)

\( \therefore \frac{dy}{dx} = y \left[ \frac{e^x}{x} + e^x \cdot \log x \right] \)

\( \implies x^{e^x} \cdot e^x \left[ \frac{1}{x} + \log x \right] \)
In simple words: This problem also uses logarithmic differentiation. After taking the natural logarithm on both sides, the product rule of differentiation is applied since the exponent \( e^x \) and \( \log x \) form a product.

🎯 Exam Tip: Remember the product rule: \( (uv)' = u'v + uv' \). Also, be careful with the derivatives of \( e^x \) and \( \log x \).

Question 3. \( y = e^{x^x} \)
Answer:
Solution:
\( y = e^{x^x} \)
\( \therefore \log y = \log e^{x^x} = x^x \log e \)
\( \therefore \log y = x^x \quad [\therefore \log e = 1] \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (x^x) \quad \ldots (1) \)
Let \( u = x^x \)
Then \( \log u = \log x^x = x \log x \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x \log x) \)

\( \implies x \frac{d}{dx} (\log x) + (\log x) \frac{d}{dx} (x) \)

\( \implies x \cdot \frac{1}{x} + (\log x) \cdot 1 \)

\( \implies 1 + \log x \)

\( \therefore \frac{du}{dx} = u(1 + \log x) \)

\( \therefore \frac{d}{dx} (x^x) = x^x(1 + \log x) \)

\( \therefore \text{ from (1),} \)
\( \frac{dy}{dx} = e^{x^x} x^x (1 + \log x) \)
In simple words: This problem involves a nested differentiation. First, we differentiate \( e^{f(x)} \) which gives \( e^{f(x)} f'(x) \). The derivative of the inner function \( x^x \) is then calculated using logarithmic differentiation, as \( x^x \) has both a variable base and a variable exponent.

🎯 Exam Tip: When dealing with \( e^{f(x)} \), its derivative is always \( e^{f(x)} \cdot f'(x) \). Recognize and apply this chain rule structure correctly.

2. Find \( \frac{dy}{dx} \) if:

Question 1. \( y = \left(1 + \frac{1}{x}\right)^x \)
Answer:
Solution:
\( y = \left(1 + \frac{1}{x}\right)^x \)
\( \therefore \log y = \log \left(1 + \frac{1}{x}\right)^x = x \log \left(1 + \frac{1}{x}\right) \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left[ x \log \left(1 + \frac{1}{x}\right) \right] \)

\( \implies x \frac{d}{dx} \left[ \log \left(1 + \frac{1}{x}\right) \right] + \left[ \log \left(1 + \frac{1}{x}\right) \right] \frac{d}{dx} (x) \)

\( \implies x \cdot \frac{1}{1 + \frac{1}{x}} \frac{d}{dx} \left(1 + \frac{1}{x}\right) + \left[ \log \left(1 + \frac{1}{x}\right) \right] \cdot 1 \)

\( \implies x \cdot \frac{1}{\frac{x+1}{x}} \left(0 - \frac{1}{x^2}\right) + \log \left(1 + \frac{1}{x}\right) \)

\( \implies x \cdot \frac{x}{x+1} \left(- \frac{1}{x^2}\right) + \log \left(1 + \frac{1}{x}\right) \)

\( \implies - \frac{1}{x+1} + \log \left(1 + \frac{1}{x}\right) \)

\( \therefore \frac{dy}{dx} = y \left[ \log \left(1 + \frac{1}{x}\right) - \frac{1}{x+1} \right] \)

\( \implies \left(1 + \frac{1}{x}\right)^x \left[ \log \left(1 + \frac{1}{x}\right) - \frac{1}{x+1} \right] \)
In simple words: This function is differentiated using logarithmic differentiation due to its variable base and exponent. The product rule is applied to \( x \log(1+\frac{1}{x}) \), and the chain rule is crucial for differentiating the \( \log(1+\frac{1}{x}) \) term.

🎯 Exam Tip: Simplify the derivative of \( (1 + \frac{1}{x}) \) carefully as \( -\frac{1}{x^2} \). Errors in this step are common.

Question 2. \( y = (2x + 5)^x \)
Answer:
Solution:
\( y = (2x+5)^x \)
\( \therefore \log y = \log (2x+5)^x = x \log (2x+5) \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [x \log (2x+5)] \)

\( \implies x \frac{d}{dx} [\log (2x+5)] + [\log (2x+5)] \frac{d}{dx} (x) \)

\( \implies x \cdot \frac{1}{2x+5} \frac{d}{dx} (2x+5) + [\log (2x+5)] \cdot 1 \)

\( \implies x \cdot \frac{1}{2x+5} (2 \cdot 1 + 0) + \log (2x+5) \)

\( \implies \frac{2x}{2x+5} + \log (2x+5) \)

\( \therefore \frac{dy}{dx} = y \left[ \frac{2x}{2x+5} + \log (2x+5) \right] \)

\( \implies (2x+5)^x \left[ \log (2x+5) + \frac{2x}{2x+5} \right] \)
In simple words: Logarithmic differentiation is applied here for a function of the form \( f(x)^{g(x)} \). The product rule is used for differentiating \( x \log(2x+5) \), and the chain rule is necessary for \( \log(2x+5) \).

🎯 Exam Tip: Clearly distinguish between \( \log(f(x)) \) and \( f(x) \log(x) \) when applying the product and chain rules.

Question 3. \( y = \sqrt[3]{\frac{3x-1}{(2x+3)(5-x)^2}} \)
Answer:
Solution:
\( y = \sqrt[3]{\frac{3x-1}{(2x+3)(5-x)^2}} \)

\( \implies y = \left( \frac{3x-1}{(2x+3)(5-x)^2} \right)^{\frac{1}{3}} \)

\( \therefore \log y = \log \left( \frac{3x-1}{(2x+3)(5-x)^2} \right)^{\frac{1}{3}} \)

\( \implies \frac{1}{3} \log \left( \frac{3x-1}{(2x+3)(5-x)^2} \right) \)

\( \implies \frac{1}{3} [\log (3x-1) - \log (2x+3) - \log (5-x)^2] \)

\( \implies \frac{1}{3} \log (3x-1) - \frac{1}{3} \log (2x+3) - \frac{2}{3} \log (5-x) \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \frac{d}{dx} [\log (3x-1)] - \frac{1}{3} \frac{d}{dx} [\log (2x+3)] - \frac{2}{3} \frac{d}{dx} [\log (5-x)] \)

\( \implies \frac{1}{3} \cdot \frac{1}{3x-1} \frac{d}{dx} (3x-1) - \frac{1}{3} \cdot \frac{1}{2x+3} \frac{d}{dx} (2x+3) - \frac{2}{3} \cdot \frac{1}{5-x} \frac{d}{dx} (5-x) \)

\( \implies \frac{1}{3(3x-1)} (3 \cdot 1 - 0) - \frac{1}{3(2x+3)} (2 \cdot 1 + 0) - \frac{2}{3(5-x)} (0 - 1) \)

\( \implies \frac{1}{3x-1} - \frac{2}{3(2x+3)} + \frac{2}{3(5-x)} \)

\( \therefore \frac{dy}{dx} = y \left[ \frac{1}{3x-1} - \frac{2}{3(2x+3)} + \frac{2}{3(5-x)} \right] \)

\( \implies \sqrt[3]{\frac{3x-1}{(2x+3)(5-x)^2}} \left[ \frac{1}{3x-1} - \frac{2}{3(2x+3)} + \frac{2}{3(5-x)} \right] \)
In simple words: This complex fractional function is simplified by taking the logarithm first, converting divisions and powers into subtractions and multiplications. Then, each logarithmic term is differentiated individually using the chain rule.

🎯 Exam Tip: Always expand the logarithmic terms completely before differentiating. This reduces complexity and minimizes errors. Remember to handle negative signs carefully when differentiating terms like \( \log(5-x) \).

3. Find \( \frac{dy}{dx} \) if:

Question 1. \( y = (\log x)^x + x^{\log x} \)
Answer:
Solution:
Let \( y = u + v \), where \( u = (\log x)^x \) and \( v = x^{\log x} \)
\( \therefore \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \quad \ldots (1) \)
Take \( u = (\log x)^x \)
\( \therefore \log u = \log (\log x)^x = x \log (\log x) \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} [x \log (\log x)] \)

\( \implies x \frac{d}{dx} [\log (\log x)] + [\log (\log x)] \frac{d}{dx} (x) \)

\( \implies x \cdot \frac{1}{\log x} \frac{d}{dx} (\log x) + [\log (\log x)] \cdot 1 \)

\( \implies x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log (\log x) \)

\( \implies \frac{1}{\log x} + \log (\log x) \)

\( \therefore \frac{du}{dx} = u \left[ \frac{1}{\log x} + \log (\log x) \right] \)

\( \implies (\log x)^x \left[ \frac{1}{\log x} + \log (\log x) \right] \quad \ldots (2) \)
Also, \( v = x^{\log x} \)
\( \therefore \log v = \log x^{\log x} = (\log x) \log x = (\log x)^2 \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx} (\log x)^2 \)

\( \implies 2(\log x) \frac{d}{dx} (\log x) \)

\( \implies 2 \log x \cdot \frac{1}{x} \)

\( \implies \frac{2 \log x}{x} \)

\( \therefore \frac{dv}{dx} = v \left[ \frac{2 \log x}{x} \right] \)

\( \implies x^{\log x} \left[ \frac{2 \log x}{x} \right] \quad \ldots (3) \)
From (1), (2) and (3), we get
\( \frac{dy}{dx} = (\log x)^x \left[ \frac{1}{\log x} + \log (\log x) \right] + x^{\log x} \left[ \frac{2 \log x}{x} \right] \)
In simple words: When the sum of two complex functions needs to be differentiated, each term is differentiated separately using logarithmic differentiation. The results are then added together. This method simplifies the process for terms with variable bases and exponents.

🎯 Exam Tip: Break down sums into individual terms and differentiate them separately. This prevents confusion and allows for focused application of differentiation rules to each part.

Question 2. \( y = x^x + a^x \)
Answer:
Solution:
Let \( u = x^x \)
Then \( \log u = \log x^x = x \log x \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x \log x) \)

\( \implies x \frac{d}{dx} (\log x) + (\log x) \frac{d}{dx} (x) \)

\( \implies x \cdot \frac{1}{x} + (\log x) \cdot 1 \)

\( \implies 1 + \log x \)

\( \therefore \frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x) \quad \ldots (1) \)
Now, \( y = u + a^x \)
\( \therefore \frac{dy}{dx} = \frac{du}{dx} + \frac{d}{dx} (a^x) \)

\( \implies x^x(1 + \log x) + a^x \log a \quad \text{[By (1)]} \)
In simple words: For a sum of functions, differentiate each term independently. For \( x^x \), use logarithmic differentiation. For \( a^x \), apply the standard derivative formula for exponential functions with a constant base.

🎯 Exam Tip: Remember the basic derivative formulas for \( a^x \) which is \( a^x \log a \) and for \( x^n \) which is \( nx^{n-1} \). Apply logarithmic differentiation only when both base and exponent are variables.

Question 3. \( y = 10^x + 10^{x^{10}} + 10^{10^x} \)
Answer:
Solution:
Let \( u = x^x \)
Then \( \log u = \log x^x = x \log x \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x \log x) \)

\( \implies x \frac{d}{dx} (\log x) + (\log x) \frac{d}{dx} (x) \)

\( \implies x \cdot \frac{1}{x} + (\log x) \cdot 1 \)

\( \implies 1 + \log x \)

\( \therefore \frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x) \quad \ldots (1) \)
Now, \( y = 10^x + 10^{x^{10}} + 10^{10^x} \)
\( \therefore \frac{dy}{dx} = \frac{d}{dx} (10^x) + \frac{d}{dx} (10^{x^{10}}) + \frac{d}{dx} (10^{10^x}) \)

\( \implies 10^x \log 10 + 10^{x^{10}} \log 10 \cdot \frac{d}{dx} (x^{10}) + 10^{10^x} \log 10 \cdot \frac{d}{dx} (10^x) \)

\( \implies 10^x \log 10 + 10^{x^{10}} \log 10 \cdot (10x^9) + 10^{10^x} \log 10 \cdot (10^x \log 10) \)

\( \implies 10^x \log 10 + 10^{x^{10}} \cdot 10x^9 \log 10 + 10^{10^x} \cdot 10^x (\log 10)^2 \)
In simple words: This problem involves differentiating a sum of three exponential terms. Each term is differentiated using the chain rule, remembering that the derivative of \( a^{f(x)} \) is \( a^{f(x)} \log a \cdot f'(x) \). This requires careful application of the chain rule to the exponents.

🎯 Exam Tip: Break down composite functions into simpler layers. For \( 10^{x^{10}} \), \( f(x) = x^{10} \), and for \( 10^{10^x} \), \( f(x) = 10^x \). Differentiate each exponent carefully.