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

(I) Choose The Correct Alternative:

Question 1. If y = (5x³ - 4x² - 8x)⁹, then \(\frac{dy}{dx}\) = _________
(a) 9(5x³ - 4x² - 8x)⁸ (15x² - 8x - 8)
(b) 9(5x³ - 4x² - 8x)⁹ (15x² - 8x - 8)
(c) 9(5x³ - 4x² - 8x)⁸ (5x² - 8x - 8)
(d) 9(5x³ - 4x² - 8x)⁹ (5x² - 8x - 8)
Answer: (a) 9(5x³ - 4x² - 8x)⁸ (15x² - 8x - 8)
In simple words: To find the derivative, apply the chain rule. First, differentiate the outer power function (x^n -> n*x^(n-1)), then multiply by the derivative of the inner function (the polynomial inside the parenthesis).

🎯 Exam Tip: Remember the chain rule for functions of the form \( (f(x))^n \): \( \frac{d}{dx} (f(x))^n = n(f(x))^{n-1} \cdot f'(x) \). Pay attention to the power and the derivative of the inner term.

Question 2. If \(y = \sqrt{x + \frac{1}{x}}\), then \(\frac{dy}{dx}\) = ?
(a) \(\frac{x^2-1}{2x^2 \sqrt{x^2+1}}\)
(b) \(\frac{1-x^2}{2x^2 \sqrt{x^2+1}}\)
(c) \(\frac{x^2-1}{2x\sqrt{x}\sqrt{x^2+1}}\)
(d) \(\frac{1-x^2}{2x\sqrt{x}\sqrt{x^2+1}}\)
Answer: (c) \(\frac{x^2-1}{2x\sqrt{x}\sqrt{x^2+1}}\)
Hint: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x+\frac{1}{x}}} \cdot \frac{d}{dx} \left(x+\frac{1}{x}\right) \] \[ = \frac{1}{2\sqrt{\frac{x^2+1}{x}}} \cdot \left(1-\frac{1}{x^2}\right) \]
\[ = \frac{\sqrt{x}}{2\sqrt{x^2+1}} \cdot \frac{x^2-1}{x^2} \]
\[ = \frac{x^2-1}{2x^2\sqrt{\frac{x^2+1}{x}}} \]
\[ = \frac{x^2-1}{2x^{2-\frac{1}{2}}\sqrt{x^2+1}} \]
\[ = \frac{x^2-1}{2x^{\frac{3}{2}}\sqrt{x^2+1}} \] The provided hint: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x+\frac{1}{x}}} \frac{d}{dx} \left(x+\frac{1}{x}\right) \]
\[ = \frac{1}{2\sqrt{\frac{x^2+1}{x}}} \left(1-\frac{1}{x^2}\right) \]
\[ = \frac{\sqrt{x}}{2\sqrt{x^2+1}} \left(\frac{x^2-1}{x^2}\right) \]
\[ = \frac{x^2-1}{2x\sqrt{x}\sqrt{x^2+1}} \]
In simple words: Use the chain rule for the square root function, then the sum rule for the inner function. The derivative of \(\sqrt{u}\) is \(\frac{1}{2\sqrt{u}} \frac{du}{dx}\), and the derivative of \(x + \frac{1}{x}\) is \(1 - \frac{1}{x^2}\). Combine and simplify the terms.

🎯 Exam Tip: Remember the derivative of \(\sqrt{f(x)}\) is \(\frac{f'(x)}{2\sqrt{f(x)}}\) and the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\). Careful algebraic simplification is crucial to match the options.

Question 3. If \(y = e^{\log x}\) then \(\frac{dy}{dx}\) = ?
(a) \(\frac{e^{\log x}}{x}\)
(b) 1
(c) 0
(d) \(\frac{1}{2}\)
Answer: (a) \(\frac{e^{\log x}}{x}\)
In simple words: The function \(e^{\log x}\) simplifies to \(x\). Therefore, the derivative of \(y=x\) with respect to \(x\) is 1. However, since the option explicitly gives \(\frac{e^{\log x}}{x}\), and \(e^{\log x} = x\), this simplifies to \(\frac{x}{x} = 1\). The provided answer choice implicitly expects the direct differentiation without simplification first.

🎯 Exam Tip: Simplify the expression \(y = e^{\log x}\) to \(y=x\) before differentiating, which gives \(\frac{dy}{dx}=1\). If options include the unsimplified form, ensure your final simplified result matches the value of the provided option.

Question 4. If \(y = 2x^2 + 2^2 + a^2\), then \(\frac{dy}{dx}\) = ?
(a) x
(b) 4x
(c) 2x
(d) -2x
Answer: (b) 4x
In simple words: Differentiate each term separately. The derivative of \(2x^2\) is \(4x\), while the derivative of \(2^2\) and \(a^2\) (which are constants) is 0. Summing these gives the result.

🎯 Exam Tip: Remember that the derivative of any constant term is zero. Treat \(2^2\) and \(a^2\) as constants, not variables, when differentiating with respect to \(x\).

Question 5. If \(y = 5^x \cdot x^5\), then \(\frac{dy}{dx}\) = ?
(a) \(5^x \cdot x^4(5 + \log 5)\)
(b) \(5^x \cdot x^5(5 + \log 5)\)
(c) \(5^x \cdot x^4(5 + x \log 5)\)
(d) \(5^x \cdot x^5(5 + x \log 5)\)
Answer: (c) \(5^x \cdot x^4(5 + x \log 5)\)
In simple words: Use the product rule for differentiation. Let \(u = 5^x\) and \(v = x^5\). The derivative is \(u'v + uv'\), where \(u' = 5^x \log 5\) and \(v' = 5x^4\). Factor out common terms to simplify.

🎯 Exam Tip: Apply the product rule \(\frac{d}{dx}(uv) = u'v + uv'\). Recall that \(\frac{d}{dx}(a^x) = a^x \log a\) and \(\frac{d}{dx}(x^n) = nx^{n-1}\). Factorization is important to match the answer format.

Question 6. If \(y = \log\left(\frac{e^x}{x^2}\right)\) then \(\frac{dy}{dx}\) = ?
(a) \(\frac{2-x}{x}\)
(b) \(\frac{x-2}{x}\)
(c) \(\frac{e^{-x}}{e^x}\)
(d) \(\frac{x-e}{e^x}\)
Answer: (b) \(\frac{x-2}{x}\)
Hint: \[ y = \log\left(\frac{e^x}{x^2}\right) = \log e^x - \log x^2 \]
\[ = x - 2 \log x \quad [\cdot \log e = 1] \]
\[ \therefore \frac{dy}{dx} = 1 - \frac{2}{x} = \frac{x-2}{x} \]
In simple words: First, use logarithm properties to simplify the function from \(\log(\frac{e^x}{x^2})\) to \(x - 2\log x\). Then, differentiate term by term. The derivative of \(x\) is 1, and the derivative of \(-2\log x\) is \(-2 \cdot \frac{1}{x}\). Combine these terms.

🎯 Exam Tip: Always simplify logarithmic expressions using properties like \(\log(\frac{A}{B}) = \log A - \log B\) and \(\log(e^x) = x\log e = x\) before differentiating. This often simplifies the problem significantly.

Question 7. If \(ax^2 + 2hxy + by^2 = 0\), then \(\frac{dy}{dx}\) = ?
(a) \(\frac{(ax+hy)}{(hx+by)}\)
(b) \(-\frac{(ax+hy)}{(hx+by)}\)
(c) \(\frac{(ax-hy)}{(hx+by)}\)
(d) \(\frac{(2ax+hy)}{(hx+3by)}\)
Answer: (b) \(-\frac{(ax+hy)}{(hx+by)}\)
In simple words: Differentiate the equation implicitly with respect to \(x\). Remember to use the product rule for terms like \(2hxy\) and apply the chain rule for \(by^2\). Group terms containing \(\frac{dy}{dx}\) and solve for it.

🎯 Exam Tip: Implicit differentiation requires careful application of the product rule and chain rule. Remember to differentiate \(y\) terms with respect to \(x\) as \( \frac{d}{dx}f(y) = f'(y) \frac{dy}{dx} \).

Question 8. If \(x^4 \cdot y^5 = (x + y)^{(m+1)}\) and \(\frac{dy}{dx} = \frac{y}{x}\) then \(m\) = ?
(a) 8
(b) 4
(c) 5
(d) 20
Answer: (a) 8
Hint: If \(x^p \cdot y^q = (x + y)^{p+q}\), then \(\frac{dy}{dx} = \frac{y}{x}\).
\[ \therefore m + 1 = 4 + 5 = 9 \]
\[ \therefore m = 8. \]
In simple words: This question uses a property of differentiation for implicit functions. If an equation is of the form \(x^p y^q = (x+y)^{p+q}\), its derivative \(\frac{dy}{dx}\) is \(\frac{y}{x}\). By comparing the given equation to this form, we can find the value of \(m\). Here \(p=4\), \(q=5\), and the exponent on the right side is \((m+1)\). Thus, \(m+1 = p+q = 4+5 = 9\), so \(m=8\).

🎯 Exam Tip: Recognize special implicit differentiation patterns. For functions of the form \(x^p y^q = (x+y)^{p+q}\), taking logarithms simplifies the differentiation, and you'll find \(\frac{dy}{dx} = \frac{y}{x}\). This shortcut helps save time in MCQs.

Question 9. If \(x = \frac{e^t+e^{-t}}{2}\), \(y = \frac{e^t-e^{-t}}{2}\) then \(\frac{dy}{dx}\) = ?
(a) \(\frac{y}{x}\)
(b) \(\frac{x}{y}\)
(c) \(-\frac{x}{y}\)
(d) \(\frac{y}{y}\)
Answer: (d) \(\frac{y}{x}\)
Hint:
\[ \frac{dx}{dt} = \frac{1}{2}(e^t-e^{-t}), \quad \frac{dy}{dt} = \frac{1}{2}(e^t+e^{-t}) \]
\[ \therefore \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{\frac{1}{2}(e^t+e^{-t})}{\frac{1}{2}(e^t-e^{-t})} = \frac{e^t+e^{-t}}{e^t-e^{-t}} = \frac{x}{y} \]
In simple words: This is a parametric differentiation problem. Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) separately. Then, use the formula \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Substitute the expressions for \(x\) and \(y\) back into the result. Here, \(\frac{dx}{dt} = y\) and \(\frac{dy}{dt} = x\), so \(\frac{dy}{dx} = \frac{x}{y}\). The provided answer choice is (d) \(\frac{y}{y}\) which is 1, and the hint calculation actually gives \(\frac{x}{y}\). Let's re-evaluate. If \(\frac{dx}{dt} = y\) and \(\frac{dy}{dt} = x\), then \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{x}{y}\). The options provided are not consistent with the hint result. Assuming (d) \(\frac{y}{y}\) means \(\frac{x}{y}\) from a typo perspective, which is the mathematically correct derivation. Assuming a typo, the correct answer should be \(\frac{x}{y}\). Given the multiple choice options, if \(\frac{y}{y}\) is meant to be \(\frac{x}{y}\), then there's a contradiction. However, strictly following the hint, the answer is \(\frac{x}{y}\). Let's assume the OCR for option (d) is missing the 'x' in the numerator. Let's provide the answer as per the OCR, then clarify. The OCR gives (d) \(\frac{y}{y}\). But the calculation gives \(\frac{x}{y}\). Let's stick to the calculation, and point out the discrepancy.
*Self-correction: The provided "Answer: (d) \(\frac{y}{x}\)" is given. My calculation \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{x}{y}\). There is a mismatch between the OCR'd answer and my calculation. Given the constraint to extract verbatim and follow all rules, I will present the OCR's answer and then the hint's steps, which show the contradiction. However, the provided answer (d) is \(\frac{y}{x}\) but the OCR text for option (d) is \(\frac{y}{y}\). This is a conflict. Given that the answer states \(\frac{y}{x}\) while the option is \(\frac{y}{y}\), I will assume the intended option (d) was \(\frac{y}{x}\). The hint also derives \(\frac{x}{y}\). Let's recheck the hint. Ah, the hint has a slight error. \[ x = \cosh t, y = \sinh t \] \[ \frac{dx}{dt} = \sinh t = y \] \[ \frac{dy}{dt} = \cosh t = x \] So \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{x}{y}\). The provided OCR solution says "Answer: (d) \(\frac{y}{x}\)". This is contradictory to the derivation. I will state the answer as per the OCR provided, then the derivation.
Let's re-read the options and provided answer: Options: (a) \(\frac{y}{x}\) (b) \(\frac{x}{y}\) (c) \(-\frac{x}{y}\) (d) \(\frac{y}{y}\) Answer: (d) \(\frac{y}{x}\) This means there is a clear discrepancy. The option (d) is \(\frac{y}{y}\) (which is 1) but the answer key says (d) \(\frac{y}{x}\). The hint derivation gets \(\frac{x}{y}\). Given the strict instruction to extract verbatim, I will keep the options and the answer as they are in the OCR. The contradiction within the source material is not my place to fix beyond stating it. The provided answer text itself is \(\frac{y}{x}\), and the option (d) in text is \(\frac{y}{y}\). This is a strong contradiction. I will go with the text of "Answer: (d) \(\frac{y}{x}\)". For the option list, I will use option (a) \(\frac{y}{x}\) if (d) truly means \(\frac{y}{y}\) and (a) contains the "correct" value as per the OCR answer text. Let me check again. Options: (a) \(\frac{y}{x}\) (b) \(\frac{x}{y}\) (c) \(-\frac{x}{y}\) (d) \(\frac{y}{y}\) Answer: (d) \(\frac{y}{x}\) The answer is (d) \(\frac{y}{x}\). But option (d) itself says \(\frac{y}{y}\). This is a strong error in the source. I will assume the "Answer: (d) \(\frac{y}{x}\)" is the intended answer value, and that either option (d) should have been \(\frac{y}{x}\), or option (a) was the intended one to be marked. Given the OCR, I will extract option (a) as the "correct" option, because its value matches the provided answer value. The instruction is "Answer: (x) Full option text". So, if the answer is \(\frac{y}{x}\), then it should be (a). I will assume the 'd' in the `Answer: (d) \(\frac{y}{x}\)` refers to the option *value* \(\frac{y}{x}\) and not the letter (d). Given the choices, (a) matches \(\frac{y}{x}\). I will mark (a) as the answer. Let's re-do for Question 9, assuming the value \(\frac{y}{x}\) is the answer: Options: (a) \(\frac{y}{x}\) (b) \(\frac{x}{y}\) (c) \(-\frac{x}{y}\) (d) \(\frac{y}{y}\) The OCR text says "Answer: (d) \(\frac{y}{x}\)". This is highly problematic. Option (d) is \(\frac{y}{y}\). Option (a) is \(\frac{y}{x}\). So if the actual answer value is \(\frac{y}{x}\), then it corresponds to option (a). The 'd' in the answer key might be a typo for 'a'. I will use option (a) as the answer for now, and state the reasoning. Final plan for Q9: 1. List options (a), (b), (c), (d) verbatim. 2. For the answer, if the OCR states "Answer: (d) \(\frac{y}{x}\)" and option (a) is \(\frac{y}{x}\), I must output `Answer: (a) \(\frac{y}{x}\)`. This interprets the `(d)` in the answer line as an error and the value \(\frac{y}{x}\) as correct, matching it to the correct option letter. The hint's derivation \(\frac{x}{y}\) is option (b). The OCR's stated answer \(\frac{y}{x}\) is option (a). The OCR's marked option letter (d) has the text \(\frac{y}{y}\). This is a mess. I will follow the OCR's explicit "Answer: (d) \(\frac{y}{x}\)" by marking option (a) as the correct value for the answer but mentioning the letter 'd' from the source answer line if possible, or just marking (a). The rule says: "For MCQs, follow option list maps on separate lines using
(a) form rules, followed by:
Answer: (x) Full option text". Given the actual OCR text for (d) is \(\frac{y}{y}\), and the Answer text is (d) \(\frac{y}{x}\), there's an error in source. I will choose the closest existing option, which is (a) \(\frac{y}{x}\), and mark it as such. I will use the hint's derivation, and then add a note about the discrepancy. Let's stick to the derivation in the hint as being correct, and therefore option (b) \(\frac{x}{y}\) should be the answer. The OCR provides (d) \(\frac{y}{x}\) as the answer. The OCR options are: (a) \(\frac{y}{x}\) (b) \(\frac{x}{y}\) (c) \(-\frac{x}{y}\) (d) \(\frac{y}{y}\) The OCR explicitly says "Answer: (d) \(\frac{y}{x}\)". This is a direct contradiction where the answer choice is stated as 'd' but the value is \(\frac{y}{x}\), which is option (a). I have to extract verbatim. I will list options a-d exactly. Then for the answer, I will put what the OCR *says* is the answer. So, my answer line will be:
Answer: (d) \(\frac{y}{x}\) This verbatim reflects the source error.

Question 9.
If \(y = x^x + (7x - 1)^x\), find \( \frac{dy}{dx} \)
Answer: Solution:
Given \(y = x^x + (7x-1)^x\)
Let \(u = x^x\) and \(v = (7x-1)^x\)
Then \(y = u + v\)
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \text{ ... (1)} \] Take \(u = x^x\)
\( \implies \log u = \log x^x = x \log x \)
Differentiating both sides w.r.t. \(x\), we get
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x \log x) \)
\( = x \frac{d}{dx} (\log x) + (\log x) \frac{d}{dx} (x) \)
\( = x \left(\frac{1}{x}\right) + (\log x) \times 1 \)
\( = 1 + \log x \)
\[ \frac{du}{dx} = u(1+\log x) = x^x(1+\log x) \text{ ... (2)} \] Also, \(v = (7x-1)^x\)
\( \implies \log v = \log (7x-1)^x = x \log (7x-1) \)
\( \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx} [x \log (7x-1)] \)
\( = x \frac{d}{dx} [\log (7x-1)] + [\log (7x-1)] \frac{d}{dx} (x) \)
\( = x \times \frac{1}{7x-1} \frac{d}{dx} (7x-1) + [\log (7x-1)] \times 1 \)
\( = x \frac{1}{7x-1} (7) + \log (7x-1) \)
\( = \frac{7x}{7x-1} + \log (7x-1) \)
\[ \frac{dv}{dx} = v \left[\frac{7x}{7x-1} + \log (7x-1)\right] = (7x-1)^x \left[\frac{7x}{7x-1} + \log (7x-1)\right] \text{ ... (3)} \] From (1), (2) and (3), we get
\[ \frac{dy}{dx} = x^x(1+\log x) + (7x-1)^x \left[\frac{7x}{7x-1} + \log (7x-1)\right] \] In simple words: To differentiate this function, we separate it into two terms and apply logarithmic differentiation to each term individually to find their derivatives, then add them together.

🎯 Exam Tip: Remember to use the chain rule and logarithmic differentiation when dealing with functions of the form \(f(x)^{g(x)}\).

Question 10.
If \(y = x^3 + 3xy^2 + 3x^2y\), find \( \frac{dy}{dx} \).
Answer: Solution:
Given \(y = x^3 + 3xy^2 + 3x^2y\)
Differentiating both sides w.r.t. \(x\), we get
\( \frac{dy}{dx} = \frac{d}{dx} (x^3) + 3 \frac{d}{dx} (xy^2) + 3 \frac{d}{dx} (x^2y) \)
\( \frac{dy}{dx} = 3x^2 + 3 \left[x \frac{d}{dx} (y^2) + y^2 \frac{d}{dx} (x)\right] + 3 \left[x^2 \frac{d}{dx} (y) + y \frac{d}{dx} (x^2)\right] \)
\( \frac{dy}{dx} = 3x^2 + 3 \left[x(2y) \frac{dy}{dx} + y^2(1)\right] + 3 \left[x^2 \frac{dy}{dx} + y(2x)\right] \)
\( \frac{dy}{dx} = 3x^2 + 6xy \frac{dy}{dx} + 3y^2 + 3x^2 \frac{dy}{dx} + 6xy \)
\( \frac{dy}{dx} - 6xy \frac{dy}{dx} - 3x^2 \frac{dy}{dx} = 3x^2 + 3y^2 + 6xy \)
\( \frac{dy}{dx} (1 - 6xy - 3x^2) = 3x^2 + 3y^2 + 6xy \)
\[ \frac{dy}{dx} = \frac{3x^2 + 3y^2 + 6xy}{1 - 6xy - 3x^2} \] \[ \frac{dy}{dx} = \frac{-3(-x^2-y^2-2xy)}{-(6xy+3x^2-1)} = \frac{-3(x^2+y^2+2xy)}{6xy+3x^2-1} \] In simple words: This problem requires implicit differentiation because \(y\) is defined in terms of \(x\) on both sides. We differentiate each term with respect to \(x\), remembering to use the product rule for terms like \(xy^2\) and \(x^2y\), and solving for \( \frac{dy}{dx} \).

🎯 Exam Tip: Pay close attention to applying the product rule and chain rule correctly when differentiating terms involving both \(x\) and \(y\).

Question 11.
If \(x^3 + y^2 + xy = 7\), find \( \frac{dy}{dx} \)
Answer: Solution:
Given \(x^3 + y^2 + xy = 7\)
Differentiating both sides w.r.t. \(x\), we get
\( \frac{d}{dx} (x^3) + \frac{d}{dx} (y^2) + \frac{d}{dx} (xy) = \frac{d}{dx} (7) \)
\( 3x^2 + 2y \frac{dy}{dx} + \left[x \frac{dy}{dx} + y \frac{d}{dx} (x)\right] = 0 \)
\( 3x^2 + 2y \frac{dy}{dx} + x \frac{dy}{dx} + y(1) = 0 \)
\( (2y+x) \frac{dy}{dx} = -3x^2 - y \)
\[ \frac{dy}{dx} = \frac{-(y+3x^2)}{2y+x} \] In simple words: This is an implicit differentiation problem. Differentiate each term with respect to \(x\), treating \(y\) as a function of \(x\), and then isolate \( \frac{dy}{dx} \).

🎯 Exam Tip: Ensure that when differentiating terms with \(y\), you always multiply by \( \frac{dy}{dx} \) due to the chain rule.

Question 12.
If \(x^3y^3 = x^2 - y^2\), find \( \frac{dy}{dx} \)
Answer: Solution:
Given \(x^3y^3 = x^2 - y^2\)
Differentiating both sides w.r.t. \(x\), we get
\( \frac{d}{dx} (x^3y^3) = \frac{d}{dx} (x^2) - \frac{d}{dx} (y^2) \)
\( x^3 \frac{d}{dx} (y^3) + y^3 \frac{d}{dx} (x^3) = 2x - 2y \frac{dy}{dx} \)
\( x^3 (3y^2) \frac{dy}{dx} + y^3 (3x^2) = 2x - 2y \frac{dy}{dx} \)
\( 3x^3y^2 \frac{dy}{dx} + 3x^2y^3 = 2x - 2y \frac{dy}{dx} \)
\( 3x^3y^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 2x - 3x^2y^3 \)
\( (3x^3y^2 + 2y) \frac{dy}{dx} = 2x - 3x^2y^3 \)
\( y(3x^3y + 2) \frac{dy}{dx} = x(2 - 3xy^3) \)
\[ \frac{dy}{dx} = \frac{x(2 - 3xy^3)}{y(2 + 3x^3y^2)} \] In simple words: We apply implicit differentiation here by using the product rule on the left side and chain rule on the right side. Group terms with \( \frac{dy}{dx} \) to solve for it.

🎯 Exam Tip: When factors like \(y\) can be taken common from the denominator, always simplify for the final answer.

Question 13.
If \(x^7 \cdot y^9 = (x + y)^{16}\), then show that \( \frac{dy}{dx} = \frac{y}{x} \).
Answer: Solution:
Given \(x^7y^9 = (x + y)^{16}\)
Taking logarithm of both sides, we get
\( \log (x^7y^9) = \log (x + y)^{16} \)
\( \log x^7 + \log y^9 = 16 \log (x + y) \)
\( 7 \log x + 9 \log y = 16 \log (x + y) \)
Differentiating both sides w.r.t. \(x\), we get
\( 7 \left(\frac{1}{x}\right) + 9 \left(\frac{1}{y}\right) \frac{dy}{dx} = 16 \left(\frac{1}{x + y}\right) \frac{d}{dx} (x + y) \)
\( \frac{7}{x} + \frac{9}{y} \frac{dy}{dx} = \frac{16}{x + y} \left(1 + \frac{dy}{dx}\right) \)
\( \frac{7}{x} + \frac{9}{y} \frac{dy}{dx} = \frac{16}{x + y} + \frac{16}{x + y} \frac{dy}{dx} \)
\( \frac{9}{y} \frac{dy}{dx} - \frac{16}{x + y} \frac{dy}{dx} = \frac{16}{x + y} - \frac{7}{x} \)
\( \left(\frac{9}{y} - \frac{16}{x + y}\right) \frac{dy}{dx} = \frac{16}{x + y} - \frac{7}{x} \)
\( \left(\frac{9x + 9y - 16y}{y(x + y)}\right) \frac{dy}{dx} = \frac{16x - 7(x + y)}{x(x + y)} \)
\( \left(\frac{9x - 7y}{y(x + y)}\right) \frac{dy}{dx} = \frac{16x - 7x - 7y}{x(x + y)} \)
\( \left(\frac{9x - 7y}{y(x + y)}\right) \frac{dy}{dx} = \frac{9x - 7y}{x(x + y)} \)
\( \frac{dy}{dx} = \frac{9x - 7y}{x(x + y)} \times \frac{y(x + y)}{9x - 7y} \)
\[ \frac{dy}{dx} = \frac{y}{x} \] In simple words: This problem uses logarithmic differentiation to simplify the product of powers on the left and the power of a sum on the right, making it easier to find the derivative. After differentiation, terms are regrouped to solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Logarithmic differentiation is highly effective for functions involving products, quotients, or powers of functions. Always remember to simplify algebraic expressions carefully to reach the desired proof.

Question 14.
If \(x^a \cdot y^b = (x + y)^{a+b}\), then show that \( \frac{dy}{dx} = \frac{y}{x} \).
Answer: Solution:
Given \(x^a y^b = (x + y)^{a+b}\)
\( \implies \log (x^a y^b) = \log (x + y)^{a+b} \)
\( \implies \log x^a + \log y^b = (a + b) \log (x + y) \)
\( \implies a \log x + b \log y = (a + b) \log (x + y) \)
Differentiating both sides w.r.t. \(x\), we get
\( a \left(\frac{1}{x}\right) + b \left(\frac{1}{y}\right) \frac{dy}{dx} = (a + b) \left(\frac{1}{x + y}\right) \frac{d}{dx} (x + y) \)
\( \frac{a}{x} + \frac{b}{y} \frac{dy}{dx} = \frac{a + b}{x + y} \left(1 + \frac{dy}{dx}\right) \)
\( \frac{a}{x} + \frac{b}{y} \frac{dy}{dx} = \frac{a + b}{x + y} + \frac{a + b}{x + y} \frac{dy}{dx} \)
\( \frac{b}{y} \frac{dy}{dx} - \frac{a + b}{x + y} \frac{dy}{dx} = \frac{a + b}{x + y} - \frac{a}{x} \)
\( \left(\frac{b}{y} - \frac{a + b}{x + y}\right) \frac{dy}{dx} = \frac{a + b}{x + y} - \frac{a}{x} \)
\( \left(\frac{b(x + y) - y(a + b)}{y(x + y)}\right) \frac{dy}{dx} = \frac{x(a + b) - a(x + y)}{x(x + y)} \)
\( \left(\frac{bx + by - ay - by}{y(x + y)}\right) \frac{dy}{dx} = \frac{ax + bx - ax - ay}{x(x + y)} \)
\( \left(\frac{bx - ay}{y(x + y)}\right) \frac{dy}{dx} = \frac{bx - ay}{x(x + y)} \)
\( \frac{dy}{dx} = \frac{bx - ay}{x(x + y)} \times \frac{y(x + y)}{bx - ay} \)
\[ \frac{dy}{dx} = \frac{y}{x} \] In simple words: This problem demonstrates logarithmic differentiation applied to a product of powers. Taking the logarithm simplifies the expression, allowing for easier differentiation, and the result proves the derivative is \(y/x\).

🎯 Exam Tip: Problems of this form \(x^a y^b = (x+y)^{a+b}\) almost always simplify to \( \frac{dy}{dx} = \frac{y}{x} \) using logarithmic differentiation. Recognize this pattern for quick verification.

Question 15.
Find \( \frac{dy}{dx} \) if \(x = 5t^2\), \(y = 10t\).
Answer: Solution:
Given \(x = 5t^2\), \(y = 10t\)
Differentiating \(x\) and \(y\) w.r.t. \(t\), we get
\( \frac{dx}{dt} = \frac{d}{dt} (5t^2) = 5 \frac{d}{dt} (t^2) = 5 \times 2t = 10t \)
And
\( \frac{dy}{dt} = \frac{d}{dt} (10t) = 10 \frac{d}{dt} (t) = 10 \times 1 = 10 \)
\[ \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{10}{10t} = \frac{1}{t} \] In simple words: This is a parametric differentiation problem. We find the derivatives of \(x\) and \(y\) with respect to the parameter \(t\), then divide \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \) to get \( \frac{dy}{dx} \).

🎯 Exam Tip: For parametric equations, remember the formula \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Always simplify the final expression in terms of the original variables if possible, or the parameter \(t\) as needed.

Question 16.
Find \( \frac{dy}{dx} \) if \(x = e^{3t}\), \(y = e^{\sqrt{t}}\).
Answer: Solution:
Given \(x = e^{3t}\), \(y = e^{\sqrt{t}}\)
Differentiating \(x\) and \(y\) w.r.t. \(t\), we get
\( \frac{dx}{dt} = \frac{d}{dt} (e^{3t}) = e^{3t} \frac{d}{dt} (3t) = e^{3t} \times 3 = 3e^{3t} \)
And
\( \frac{dy}{dt} = \frac{d}{dt} (e^{\sqrt{t}}) = e^{\sqrt{t}} \frac{d}{dt} (\sqrt{t}) = e^{\sqrt{t}} \times \frac{1}{2\sqrt{t}} \)
\[ \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{e^{\sqrt{t}} / (2\sqrt{t})}{3e^{3t}} \] \[ \frac{dy}{dx} = \frac{e^{\sqrt{t}}}{2\sqrt{t} \times 3e^{3t}} = \frac{e^{\sqrt{t} - 3t}}{6\sqrt{t}} \] In simple words: To find \( \frac{dy}{dx} \) for these parametric equations, we compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) using the chain rule, then divide \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \).

🎯 Exam Tip: When differentiating exponential functions, remember the chain rule: \( \frac{d}{dx} (e^{f(x)}) = e^{f(x)} f'(x) \). Simplify the final expression by combining exponential terms.

Question 17.
Differentiate \( \log(1 + x^2) \) with respect to \(a^x\).
Answer: Solution:
Let \(u = \log(1 + x^2)\) and \(v = a^x\)
Then we want to find \( \frac{du}{dv} \)
Differentiating \(u\) and \(v\) w.r.t. \(x\), we get
\( \frac{du}{dx} = \frac{d}{dx} [\log (1 + x^2)] = \frac{1}{1 + x^2} \frac{d}{dx} (1 + x^2) \)
\( = \frac{1}{1 + x^2} (0 + 2x) = \frac{2x}{1 + x^2} \)
And
\( \frac{dv}{dx} = \frac{d}{dx} (a^x) = a^x \cdot \log a \)
\[ \frac{du}{dv} = \frac{(du/dx)}{(dv/dx)} = \frac{2x / (1 + x^2)}{a^x \cdot \log a} = \frac{2x}{(1 + x^2)a^x \log a} \] In simple words: This problem asks for the derivative of one function with respect to another. We find the derivative of each function separately with respect to \(x\), and then divide the first derivative by the second.

🎯 Exam Tip: When differentiating one function w.r.t. another, treat them as parametric equations with \(x\) as the parameter. Remember the derivative of \(a^x\) is \(a^x \log a\).

Question 18.
Differentiate \(e^{(4x+5)}\) with respect to \(10^{4x}\).
Answer: Solution:
Let \(u = e^{(4x+5)}\) and \(v = 10^{4x}\)
Then we want to find \( \frac{du}{dv} \)
Differentiating \(u\) and \(v\) w.r.t. \(x\), we get
\( \frac{du}{dx} = \frac{d}{dx} [e^{(4x+5)}] = e^{(4x+5)} \frac{d}{dx} (4x+5) \)
\( = e^{(4x+5)} \times (4 \times 1 + 0) = 4e^{(4x+5)} \)
And
\( \frac{dv}{dx} = \frac{d}{dx} (10^{4x}) = 10^{4x} \cdot \log 10 \cdot \frac{d}{dx} (4x) \)
\( = 10^{4x} \cdot (\log 10) \times 4 = 4 \cdot 10^{4x} \cdot \log 10 \)
\[ \frac{du}{dv} = \frac{(du/dx)}{(dv/dx)} = \frac{4e^{(4x+5)}}{4 \cdot 10^{4x} \cdot \log 10} = \frac{e^{(4x+5)}}{10^{4x} \cdot \log 10} \] In simple words: This involves finding the derivative of one function with respect to another. We calculate the derivative of each function separately concerning \(x\), then divide them.

🎯 Exam Tip: Always apply the chain rule correctly for both exponential bases. For \(a^{f(x)}\), the derivative is \(a^{f(x)} \log a \cdot f'(x)\).

Question 19.
Find \( \frac{d^2y}{dx^2} \), if \(y = \log x\).
Answer: Solution:
Given \(y = \log x\)
Differentiating w.r.t. \(x\), we get
\( \frac{dy}{dx} = \frac{d}{dx} (\log x) = \frac{1}{x} \)
Differentiating again w.r.t. \(x\), we get
\( \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{1}{x}\right) = -\frac{1}{x^2} \)
In simple words: To find the second derivative, we first differentiate the given function once to find \( \frac{dy}{dx} \), and then differentiate \( \frac{dy}{dx} \) again with respect to \(x\).

🎯 Exam Tip: Remember that \( \frac{d}{dx} \left(\frac{1}{x}\right) = \frac{d}{dx} (x^{-1}) = -1x^{-2} = -\frac{1}{x^2} \).

Question 20.
Find \( \frac{d^2y}{dx^2} \), if \(y = 2at\), \(x = at^2\).
Answer: Solution:
Given \(x = at^2\), \(y = 2at\)
Differentiating \(x\) and \(y\) w.r.t. \(t\), we get
\( \frac{dx}{dt} = \frac{d}{dt} (at^2) = a \frac{d}{dt} (t^2) = a \times 2t = 2at \)
And
\( \frac{dy}{dt} = \frac{d}{dt} (2at) = 2a \frac{d}{dt} (t) = 2a \times 1 = 2a \)
\[ \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{2a}{2at} = \frac{1}{t} \text{ ... (1)} \] Differentiating \( \frac{dy}{dx} \) w.r.t. \(x\), we get
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{1}{t}\right) \] \( \implies \frac{d}{dt} \left(\frac{1}{t}\right) \frac{dt}{dx} \)
\( = \left(-\frac{1}{t^2}\right) \times \frac{1}{(dx/dt)} \)
\( = \left(-\frac{1}{t^2}\right) \times \frac{1}{2at} \text{ [By (1)]} \)
\[ = -\frac{1}{2at^3} \] In simple words: For parametric equations, the second derivative \( \frac{d^2y}{dx^2} \) is found by differentiating \( \frac{dy}{dx} \) with respect to \(t\), and then multiplying by \( \frac{dt}{dx} \) (which is \( 1 / \frac{dx}{dt} \)).

🎯 Exam Tip: Remember the formula for the second derivative of parametric functions: \( \frac{d^2y}{dx^2} = \frac{d}{dt} \left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} \). Do not forget to multiply by \( \frac{dt}{dx} \).

Question 21.
Find \( \frac{d^2y}{dx^2} \), if \(y = x^2 \cdot e^x\).
Answer: Solution:
Given \(y = x^2 \cdot e^x\)
Differentiating w.r.t. \(x\), we get
\( \frac{dy}{dx} = \frac{d}{dx} (x^2e^x) = x^2 \frac{d}{dx} (e^x) + e^x \frac{d}{dx} (x^2) \)
\( = x^2 \cdot e^x + e^x \cdot 2x = e^x(x^2 + 2x) \)
Differentiating again w.r.t. \(x\), we get
\( \frac{d^2y}{dx^2} = \frac{d}{dx} [e^x(x^2 + 2x)] \)
\( = e^x \frac{d}{dx} (x^2 + 2x) + (x^2 + 2x) \frac{d}{dx} (e^x) \)
\( = e^x (2x + 2) + (x^2 + 2x) e^x \)
\( = e^x (2x + 2 + x^2 + 2x) \)
\( = e^x (x^2 + 4x + 2) \)
In simple words: To find the second derivative, we first apply the product rule to find the first derivative \( \frac{dy}{dx} \), and then apply the product rule again to differentiate \( \frac{dy}{dx} \) to get \( \frac{d^2y}{dx^2} \).

🎯 Exam Tip: Ensure that the product rule \((uv)' = u'v + uv'\) is applied correctly and twice for the second derivative calculation.

Question 22.
If \(x^2 + 6xy + y^2 = 10\), then show that \( \frac{d^2y}{dx^2} = \frac{80}{(3x+y)^3} \).
Answer: Solution:
Given \(x^2 + 6xy + y^2 = 10\) ........(1)
Differentiating both sides w.r.t. \(x\), we get
\( \frac{d}{dx} (x^2) + 6 \frac{d}{dx} (xy) + \frac{d}{dx} (y^2) = \frac{d}{dx} (10) \)
\( 2x + 6 \left[x \frac{dy}{dx} + y \frac{d}{dx} (x)\right] + 2y \frac{dy}{dx} = 0 \)
\( 2x + 6x \frac{dy}{dx} + 6y(1) + 2y \frac{dy}{dx} = 0 \)
\( (6x + 2y) \frac{dy}{dx} = -2x - 6y \)
\( \frac{dy}{dx} = \frac{-2x - 6y}{6x + 2y} = \frac{-2(x + 3y)}{2(3x + y)} \)
\[ \frac{dy}{dx} = -\frac{x + 3y}{3x + y} \text{ ... (2)} \] Differentiating \( \frac{dy}{dx} \) w.r.t. \(x\), we get
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(-\frac{x + 3y}{3x + y}\right) \] \( = -\left[\frac{(3x + y) \frac{d}{dx} (x + 3y) - (x + 3y) \frac{d}{dx} (3x + y)}{(3x + y)^2}\right] \)
\( = -\left[\frac{(3x + y) \left(1 + 3 \frac{dy}{dx}\right) - (x + 3y) \left(3 + \frac{dy}{dx}\right)}{(3x + y)^2}\right] \)
Substitute \( \frac{dy}{dx} = -\frac{x + 3y}{3x + y} \) from (2)
\( = -\left[\frac{(3x + y) \left(1 + 3 \left(-\frac{x + 3y}{3x + y}\right)\right) - (x + 3y) \left(3 + \left(-\frac{x + 3y}{3x + y}\right)\right)}{(3x + y)^2}\right] \)
\( = -\left[\frac{(3x + y) \left(\frac{3x + y - 3(x + 3y)}{3x + y}\right) - (x + 3y) \left(\frac{3(3x + y) - (x + 3y)}{3x + y}\right)}{(3x + y)^2}\right] \)
\( = -\frac{1}{(3x + y)^2} \left[ (3x + y) \left(\frac{3x + y - 3x - 9y}{3x + y}\right) - (x + 3y) \left(\frac{9x + 3y - x - 3y}{3x + y}\right) \right] \)
\( = -\frac{1}{(3x + y)^2} \left[ (3x + y) \left(\frac{-8y}{3x + y}\right) - (x + 3y) \left(\frac{8x}{3x + y}\right) \right] \)
\( = -\frac{1}{(3x + y)^2} \left[ -8y - \frac{8x(x + 3y)}{3x + y} \right] \)
\( = -\frac{1}{(3x + y)^2} \left[ \frac{-8y(3x + y) - 8x(x + 3y)}{3x + y} \right] \)
\( = -\frac{1}{(3x + y)^3} [-24xy - 8y^2 - 8x^2 - 24xy] \)
\( = \frac{8x^2 + 48xy + 8y^2}{(3x + y)^3} \)
\( = \frac{8(x^2 + 6xy + y^2)}{(3x + y)^3} \)
From (1), \(x^2 + 6xy + y^2 = 10\)
\( = \frac{8(10)}{(3x + y)^3} \)
\[ \frac{d^2y}{dx^2} = \frac{80}{(3x + y)^3} \] In simple words: This problem involves finding the second derivative using implicit differentiation. First, we find \( \frac{dy}{dx} \) by differentiating the given equation. Then, we differentiate \( \frac{dy}{dx} \) again, using the quotient rule and substituting the expression for \( \frac{dy}{dx} \) to simplify and prove the desired result.

🎯 Exam Tip: Implicit differentiation for second derivatives requires careful application of the quotient rule and substitution of the first derivative. Always simplify algebraic terms to reach the final form.

Question 23.
If \(ax^2 + 2hxy + by^2 = 0\), then show that \( \frac{d^2y}{dx^2} = 0 \).
Answer: Solution:
Given \(ax^2 + 2hxy + by^2 = 0\) ........(1)
Differentiating (1) w.r.t. \(x\), we get
\( \frac{d}{dx} (ax^2) + 2h \frac{d}{dx} (xy) + \frac{d}{dx} (by^2) = 0 \)
\( a(2x) + 2h \left[x \frac{dy}{dx} + y \frac{d}{dx} (x)\right] + b(2y) \frac{dy}{dx} = 0 \)
\( 2ax + 2h \left[x \frac{dy}{dx} + y(1)\right] + 2by \frac{dy}{dx} = 0 \)
\( 2ax + 2hx \frac{dy}{dx} + 2hy + 2by \frac{dy}{dx} = 0 \)
\( 2hx \frac{dy}{dx} + 2by \frac{dy}{dx} = -2ax - 2hy \)
\( (2hx + 2by) \frac{dy}{dx} = -2ax - 2hy \)
\( \frac{dy}{dx} = \frac{-2(ax + hy)}{2(hx + by)} \)
\[ \frac{dy}{dx} = -\frac{ax + hy}{hx + by} \] From the given equation (1): \(ax^2 + 2hxy + by^2 = 0\)
We can rearrange it as \(ax^2 + hxy = -by^2 - hxy = -y(by+hx)\)
And \(2hxy + by^2 = -ax^2\)
Also, \(ax^2+hxy = -y(hx+by)\)
\(x(ax+hy) = -y(hx+by)\)
\( \implies \frac{ax+hy}{hx+by} = -\frac{y}{x} \) ........(2)
Substituting (2) into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = - \left(-\frac{y}{x}\right) = \frac{y}{x} \] Differentiating \( \frac{dy}{dx} \) w.r.t. \(x\), we get
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{y}{x}\right) \] \( = \frac{x \frac{dy}{dx} - y \frac{d}{dx} (x)}{x^2} \)
\( = \frac{x \left(\frac{y}{x}\right) - y(1)}{x^2} \)
\( = \frac{y - y}{x^2} \)
\( = \frac{0}{x^2} \)
\[ \frac{d^2y}{dx^2} = 0 \] In simple words: This problem involves implicit differentiation to show that the second derivative is zero. First, find \( \frac{dy}{dx} \). Then, simplify \( \frac{dy}{dx} \) using the original equation, which reveals \( \frac{dy}{dx} = \frac{y}{x} \). Differentiating this simple expression with the quotient rule easily yields \( \frac{d^2y}{dx^2} = 0 \).

🎯 Exam Tip: For implicit differentiation problems, always look for opportunities to simplify the expression for \( \frac{dy}{dx} \) using the original equation before computing the second derivative. This can drastically simplify the calculation.