Maharashtra Board Class 12 Maths Part 1 Chapter 3 Differentiation 3.1 Solutions

Question 1. Find \(\frac{dy}{dx}\) if, \(y = \sqrt{x + \frac{1}{x}}\)
Answer: Given: \(y = \sqrt{x + \frac{1}{x}}\) Let \(u = x + \frac{1}{x}\) Then \(y = \sqrt{u}\) \(\therefore \frac{dy}{du} = \frac{d}{du}(u^{\frac{1}{2}}) = \frac{1}{2}u^{\frac{1}{2}-1} = \frac{1}{2}u^{-\frac{1}{2}}\) \( = \frac{1}{2\sqrt{u}}\) \( = \frac{1}{2\sqrt{x + \frac{1}{x}}}\) And \(\frac{du}{dx} = \frac{d}{dx}\left(x + \frac{1}{x}\right)\) \( = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})\) \( = 1 + (-1)x^{-2} = 1 - \frac{1}{x^2}\) \(\therefore \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\( \implies \frac{1}{2\sqrt{x + \frac{1}{x}}} \left(1 - \frac{1}{x^2}\right)\)
\( \implies \frac{1}{2}\left(x + \frac{1}{x}\right)^{-\frac{1}{2}} \left(1 - \frac{1}{x^2}\right)\) In simple words: This problem involves differentiating a composite function. We use the chain rule by substituting the inner function \(x + \frac{1}{x}\) with \(u\), differentiating \(y\) with respect to \(u\), and then \(u\) with respect to \(x\), finally multiplying the results.

🎯 Exam Tip: Remember to apply the chain rule correctly for functions of the form \(f(g(x))\). Pay close attention to power rule and derivative of \(1/x\).

Question 2. Find \(\frac{dy}{dx}\) if, \(y = \sqrt[3]{a^2 + x^2}\)
Answer: Given: \(y = \sqrt[3]{a^2 + x^2}\) Let \(u = a^2 + x^2\) Then \(y = \sqrt[3]{u}\) \(\therefore \frac{dy}{du} = \frac{d}{du}(u^{\frac{1}{3}}) = \frac{1}{3}u^{\frac{1}{3}-1} = \frac{1}{3}u^{-\frac{2}{3}}\) \( = \frac{1}{3}(a^2 + x^2)^{-\frac{2}{3}}\) And \(\frac{du}{dx} = \frac{d}{dx}(a^2 + x^2)\) \( = 0 + 2x = 2x\) \(\therefore \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\( \implies \frac{1}{3}(a^2 + x^2)^{-\frac{2}{3}} \cdot 2x\)`
\( \implies \frac{2x}{3(a^2 + x^2)^{\frac{2}{3}}}\) In simple words: This problem uses the chain rule for a cube root function. We define \(u\) as the expression inside the cube root, differentiate \(y\) with respect to \(u\), and then \(u\) with respect to \(x\), multiplying the derivatives.

🎯 Exam Tip: Convert radical forms to fractional exponents (e.g., \(\sqrt[n]{x} = x^{\frac{1}{n}}\)) before differentiating. Remember the derivative of a constant is zero.

Question 3. Find \(\frac{dy}{dx}\) if, \(y = (5x^3 - 4x^2 - 8x)^9\)
Answer: Given: \(y = (5x^3 - 4x^2 - 8x)^9\) Let \(u = 5x^3 - 4x^2 - 8x\) Then \(y = u^9\) \(\therefore \frac{dy}{du} = \frac{d}{du}(u^9) = 9u^8\) \( = 9(5x^3 - 4x^2 - 8x)^8\) And \(\frac{du}{dx} = \frac{d}{dx}(5x^3 - 4x^2 - 8x)\) \( = 5\frac{d}{dx}(x^3) - 4\frac{d}{dx}(x^2) - 8\frac{d}{dx}(x)\) \( = 5 \times 3x^2 - 4 \times 2x - 8 \times 1\) \( = 15x^2 - 8x - 8\) \(\therefore \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\( \implies 9(5x^3 - 4x^2 - 8x)^8 (15x^2 - 8x - 8)\) In simple words: This problem involves the chain rule for a power of a function. We substitute the inner polynomial with \(u\), differentiate \(y\) with respect to \(u\) using the power rule, and then differentiate \(u\) with respect to \(x\), multiplying the two derivatives together.

🎯 Exam Tip: For functions like \((f(x))^n\), the chain rule is \(n(f(x))^{n-1} \cdot f'(x)\). Be meticulous with polynomial differentiation steps.

Question 1. Find \(\frac{dy}{dx}\) if, \(y = \log(\log x)\)
Answer: Given \(y = \log(\log x)\) Let \(u = \log x\) Then \(y = \log u\) \(\therefore \frac{dy}{du} = \frac{d}{du}(\log u)\)
\( \implies = \frac{1}{u}\)
\( \implies = \frac{1}{\log x}\) And \(\frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}\) \(\therefore \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\( \implies \frac{1}{\log x} \cdot \frac{1}{x}\)
\( \implies = \frac{1}{x \log x}\) In simple words: This problem uses the chain rule for a logarithm of a logarithm. We let \(u = \log x\), differentiate \(y\) with respect to \(u\), and then \(u\) with respect to \(x\), multiplying these derivatives to find the final result.

🎯 Exam Tip: Remember the derivative of \(\log x\) is \(1/x\). Apply the chain rule carefully when dealing with nested logarithmic functions.

Question 2. Find \(\frac{dy}{dx}\) if, \(y = \log(10x^4 + 5x^3 - 3x^2 + 2)\)
Answer: Given \(y = \log(10x^4 + 5x^3 - 3x^2 + 2)\) Let \(u = 10x^4 + 5x^3 - 3x^2 + 2\) Then \(y = \log u\) \(\therefore \frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u}\) \( = \frac{1}{10x^4 + 5x^3 - 3x^2 + 2}\) And \(\frac{du}{dx} = \frac{d}{dx}(10x^4 + 5x^3 - 3x^2 + 2)\) \( = 10\frac{d}{dx}(x^4) + 5\frac{d}{dx}(x^3) - 3\frac{d}{dx}(x^2) + \frac{d}{dx}(2)\) \( = 10 \times 4x^3 + 5 \times 3x^2 - 3 \times 2x + 0\) \( = 40x^3 + 15x^2 - 6x\) \(\therefore \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\( \implies \frac{1}{10x^4 + 5x^3 - 3x^2 + 2} \times (40x^3 + 15x^2 - 6x)\)
\( \implies \frac{40x^3 + 15x^2 - 6x}{10x^4 + 5x^3 - 3x^2 + 2}\) In simple words: This problem applies the chain rule to a logarithmic function of a polynomial. We define the polynomial as \(u\), differentiate \(\log u\) with respect to \(u\) to get \(1/u\), and then differentiate \(u\) with respect to \(x\), multiplying these two results.

🎯 Exam Tip: When differentiating \(\log(f(x))\), the derivative is \(f'(x)/f(x)\). Ensure all terms in the polynomial are differentiated correctly with respect to \(x\).

Question 3. Find \(\frac{dy}{dx}\) if, \(y = \log(ax^2 + bx + c)\)
Answer: Given \(y = \log(ax^2 + bx + c)\) Let \(u = ax^2 + bx + c\) Then \(y = \log u\) \(\therefore \frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u}\) \( = \frac{1}{ax^2 + bx + c}\) And \(\frac{du}{dx} = \frac{d}{dx}(ax^2 + bx + c)\) \( = a\frac{d}{dx}(x^2) + b\frac{d}{dx}(x) + \frac{d}{dx}(c)\) \( = a \times 2x + b \times 1 + 0 = 2ax + b\) \(\therefore \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\( \implies \frac{1}{ax^2 + bx + c} \times (2ax + b)\)
\( \implies = \frac{2ax + b}{ax^2 + bx + c}\) In simple words: This problem uses the chain rule for a logarithm of a quadratic expression. We substitute the quadratic \(ax^2 + bx + c\) with \(u\), find the derivative of \(\log u\) with respect to \(u\), and then the derivative of \(u\) with respect to \(x\), combining them by multiplication.

🎯 Exam Tip: The derivative of \(\log(f(x))\) is \(f'(x)/f(x)\). For polynomials, apply the power rule to each term, remembering constants differentiate to zero.

Question 1. Find \(\frac{dy}{dx}\) if, \(y = e^{5x^2-2x+4}\)
Answer: Given: \(y = e^{5x^2-2x+4}\) Let \(u = 5x^2 - 2x + 4\) Then \(y = e^u\) \(\therefore \frac{dy}{du} = \frac{d}{du}(e^u) = e^u\) \( = e^{5x^2-2x+4}\) And \(\frac{du}{dx} = \frac{d}{dx}(5x^2 - 2x + 4)\) \( = 5\frac{d}{dx}(x^2) - 2\frac{d}{dx}(x) + \frac{d}{dx}(4)\) \( = 5 \times 2x - 2 \times 1 + 0 = 10x - 2\) \(\therefore \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\( \implies e^{5x^2-2x+4} \times (10x - 2)\)
\( \implies (10x - 2)e^{5x^2-2x+4}\) In simple words: This problem uses the chain rule for an exponential function. We set the exponent \(5x^2 - 2x + 4\) as \(u\), differentiate \(e^u\) with respect to \(u\), and then \(u\) with respect to \(x\), multiplying the two derivatives.

🎯 Exam Tip: The derivative of \(e^{f(x)}\) is \(e^{f(x)} \cdot f'(x)\). Ensure the derivative of the exponent is calculated correctly.

Question 2. Find \(\frac{dy}{dx}\) if, \(y = a^{(1+\log x)}\)
Answer: Given: \(y = a^{(1+\log x)}\) Let \(u = 1 + \log x\) Then \(y = a^u\) \(\therefore \frac{dy}{du} = \frac{d}{du}(a^u) = a^u \cdot \log a\)` \( = a^{(1+\log x)} \cdot \log a\)` And \(\frac{du}{dx} = \frac{d}{dx}(1 + \log x)\)` \( = 0 + \frac{1}{x} = \frac{1}{x}\)` \(\therefore \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)`
\( \implies a^{(1+\log x)} \cdot \log a \cdot \frac{1}{x}\) In simple words: This problem involves the chain rule for a general exponential function \(a^f(x)\). We substitute the exponent \(1 + \log x\) with \(u\), differentiate \(a^u\) with respect to \(u\), and then \(u\) with respect to \(x\), multiplying the results.

🎯 Exam Tip: Recall that the derivative of \(a^x\) is \(a^x \log a\). When using the chain rule, ensure the derivative of the exponent is correctly factored in.

Question 3. Find \(\frac{dy}{dx}\) if, \(y = 5^{(x+\log x)}\)
Answer: Given: \(y = 5^{(x+\log x)}\) Let \(u = x + \log x\) Then \(y = 5^u\) \(\therefore \frac{dy}{du} = \frac{d}{du}(5^u) = 5^u \cdot \log 5\) \( = 5^{(x+\log x)} \cdot \log 5\) And \(\frac{du}{dx} = \frac{d}{dx}(x + \log x)\) \( = 1 + \frac{1}{x}\) \(\therefore \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\( \implies 5^{(x+\log x)} \cdot \log 5 \cdot \left(1 + \frac{1}{x}\right)\) In simple words: This problem applies the chain rule to an exponential function with a base of 5. We define the exponent \(x + \log x\) as \(u\), find the derivative of \(5^u\) with respect to \(u\), and then the derivative of \(u\) with respect to \(x\), multiplying these two results.

🎯 Exam Tip: For functions of the form \(b^{f(x)}\), the derivative is \(b^{f(x)} \log b \cdot f'(x)\). Accurately differentiate the exponent, which here is a sum of two functions.