OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Exercise 8 (H)

S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(h)

Find \( \frac { dy }{ dx } \) if

Question 1. \( x^2 + y^2 = a^2 \)
Answer: Given the equation \( x^2 + y^2 = a^2 \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(a^2) \)
\( 2x + 2y \frac{dy}{dx} = 0 \)
\( \implies 2y \frac{dy}{dx} = -2x \)
\( \implies \frac{dy}{dx} = \frac{-2x}{2y} \)
\( \implies \frac{dy}{dx} = -\frac{x}{y} \) To solve this, we apply implicit differentiation, treating \(y\) as a function of \(x\).
In simple words: When you have an equation with both \(x\) and \(y\), and you want to find how \(y\) changes as \(x\) changes, you differentiate each part. Remember to use the chain rule for terms involving \(y\).

🎯 Exam Tip: When differentiating terms with \(y^n\), remember to apply the chain rule: \( \frac{d}{dx}(y^n) = ny^{n-1} \frac{dy}{dx} \).

Question 2. \( y^2 = 4ax \)
Answer: Given the equation \( y^2 = 4ax \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) \)
\( 2y \frac{dy}{dx} = 4a \)
\( \implies \frac{dy}{dx} = \frac{4a}{2y} \)
\( \implies \frac{dy}{dx} = \frac{2a}{y} \) This is the slope of the tangent line to the parabola at any point \( (x, y) \).
In simple words: Take the derivative of each side of the equation. For the \( y^2 \) term, you get \( 2y \) times \( \frac{dy}{dx} \). For \( 4ax \), you just get \( 4a \). Then, rearrange the equation to find what \( \frac{dy}{dx} \) is equal to.

🎯 Exam Tip: Constants like \( 4a \) should be treated carefully; they multiply the derivative of the variable part. The derivative of a constant alone is zero.

Question 3. \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
Answer: Given the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{1}{a^2} \frac{d}{dx}(x^2) + \frac{1}{b^2} \frac{d}{dx}(y^2) = \frac{d}{dx}(1) \)
\( \frac{1}{a^2} (2x) + \frac{1}{b^2} (2y \frac{dy}{dx}) = 0 \)
\( \implies \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \)
\( \implies \frac{2y}{b^2} \frac{dy}{dx} = -\frac{2x}{a^2} \)
\( \implies \frac{dy}{dx} = -\frac{2x}{a^2} \cdot \frac{b^2}{2y} \)
\( \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y} \) This process is crucial for finding the slope of tangent lines for ellipses and other implicitly defined curves.
In simple words: This equation is for an ellipse. To find \( \frac{dy}{dx} \), differentiate each fraction. For \( \frac{x^2}{a^2} \), you get \( \frac{2x}{a^2} \). For \( \frac{y^2}{b^2} \), you get \( \frac{2y}{b^2} \) multiplied by \( \frac{dy}{dx} \). The right side, being a constant, differentiates to zero. Then, solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Remember that \( a^2 \) and \( b^2 \) are constants, so they remain in the denominator when differentiating. The derivative of a constant like 1 is always 0.

Question 4. \( xy = c^2 \)
Answer: Given the equation \( xy = c^2 \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \), using the product rule for \( xy \):
\( \frac{d}{dx}(x \cdot y) = \frac{d}{dx}(c^2) \)
\( x \frac{dy}{dx} + y \cdot 1 = 0 \)
\( \implies x \frac{dy}{dx} + y = 0 \)
\( \implies x \frac{dy}{dx} = -y \)
\( \implies \frac{dy}{dx} = -\frac{y}{x} \) The product rule is essential here because both \(x\) and \(y\) are variables related by the equation.
In simple words: Since \( x \) and \( y \) are multiplied together, you must use the product rule. This means taking the derivative of the first term times the second, plus the first term times the derivative of the second. The right side is a constant, so its derivative is zero. Then, arrange the terms to find \( \frac{dy}{dx} \).

🎯 Exam Tip: The product rule states: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \). Make sure to apply it correctly when variables are multiplied.

Question 5. \( x^3 + 8xy + y^3 = 64 \)
Answer: Given the equation \( x^3 + 8xy + y^3 = 64 \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(x^3) + \frac{d}{dx}(8xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(64) \)
\( 3x^2 + 8\left(x \frac{dy}{dx} + y \cdot 1\right) + 3y^2 \frac{dy}{dx} = 0 \)
\( \implies 3x^2 + 8x \frac{dy}{dx} + 8y + 3y^2 \frac{dy}{dx} = 0 \)
\( \implies (8x + 3y^2) \frac{dy}{dx} = -3x^2 - 8y \)
\( \implies \frac{dy}{dx} = -\frac{3x^2 + 8y}{8x + 3y^2} \) This method of implicit differentiation helps us find slopes even for complex curves.
In simple words: Differentiate each part of the equation. Remember to use the product rule for \( 8xy \) and the chain rule for \( y^3 \). Group all terms with \( \frac{dy}{dx} \) on one side and the other terms on the other side. Then, solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Always pay attention to the signs when moving terms across the equality. A common mistake is forgetting to change the sign.

Question 6. \( x^3 + y^3 = 3axy \)
Answer: Given the equation \( x^3 + y^3 = 3axy \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(3axy) \)
\( 3x^2 + 3y^2 \frac{dy}{dx} = 3a\left(x \frac{dy}{dx} + y \cdot 1\right) \)
\( \implies 3x^2 + 3y^2 \frac{dy}{dx} = 3ax \frac{dy}{dx} + 3ay \)
\( \implies 3y^2 \frac{dy}{dx} - 3ax \frac{dy}{dx} = 3ay - 3x^2 \)
\( \implies (3y^2 - 3ax) \frac{dy}{dx} = 3ay - 3x^2 \)
\( \implies \frac{dy}{dx} = \frac{3ay - 3x^2}{3y^2 - 3ax} \)
\( \implies \frac{dy}{dx} = \frac{3(ay - x^2)}{3(y^2 - ax)} \)
\( \implies \frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax} \) This type of differentiation is fundamental for understanding curves in coordinate geometry.
In simple words: Differentiate all terms. Use the product rule for \( 3axy \) because \( x \) and \( y \) are multiplied. Group all the \( \frac{dy}{dx} \) terms together. Factor out \( \frac{dy}{dx} \) and then divide to solve for it. Make sure to simplify the fraction if possible.

🎯 Exam Tip: When all terms in the numerator and denominator share a common factor (like 3 in this case), always simplify the fraction for the final answer.

Question 7. \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \)
Answer: Given the equation \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy) + \frac{d}{dx}(by^2) + \frac{d}{dx}(2gx) + \frac{d}{dx}(2fy) + \frac{d}{dx}(c) = \frac{d}{dx}(0) \)
\( 2ax + 2h\left(x \frac{dy}{dx} + y \cdot 1\right) + 2by \frac{dy}{dx} + 2g + 2f \frac{dy}{dx} + 0 = 0 \)
\( \implies 2ax + 2hx \frac{dy}{dx} + 2hy + 2by \frac{dy}{dx} + 2g + 2f \frac{dy}{dx} = 0 \)
\( \implies (2hx + 2by + 2f) \frac{dy}{dx} = -2ax - 2hy - 2g \)
\( \implies 2(hx + by + f) \frac{dy}{dx} = -2(ax + hy + g) \)
\( \implies \frac{dy}{dx} = -\frac{2(ax + hy + g)}{2(hx + by + f)} \)
\( \implies \frac{dy}{dx} = -\frac{ax + hy + g}{hx + by + f} \) This general form is common in coordinate geometry and helps describe conic sections.
In simple words: This is a long equation, but the steps are the same. Differentiate each term carefully. Use the product rule for \( 2hxy \). Group all terms that have \( \frac{dy}{dx} \) on one side, and move all other terms to the opposite side. Then, factor out \( \frac{dy}{dx} \) and divide to find its value.

🎯 Exam Tip: Remember that terms like \( 2gx \) differentiate to \( 2g \) and \( 2fy \) differentiate to \( 2f \frac{dy}{dx} \). Constants like \( c \) always differentiate to zero.

Question 8. \( (x^2 + y^2)^2 = xy \)
Answer: Given the equation \( (x^2 + y^2)^2 = xy \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}((x^2 + y^2)^2) = \frac{d}{dx}(xy) \)
Using the chain rule on the left side and product rule on the right side:
\( 2(x^2 + y^2) \cdot \frac{d}{dx}(x^2 + y^2) = x \frac{dy}{dx} + y \cdot 1 \)
\( 2(x^2 + y^2) (2x + 2y \frac{dy}{dx}) = x \frac{dy}{dx} + y \)
\( \implies 4x(x^2 + y^2) + 4y(x^2 + y^2) \frac{dy}{dx} = x \frac{dy}{dx} + y \)
\( \implies 4y(x^2 + y^2) \frac{dy}{dx} - x \frac{dy}{dx} = y - 4x(x^2 + y^2) \)
\( \implies (4y(x^2 + y^2) - x) \frac{dy}{dx} = y - 4x(x^2 + y^2) \)
\( \implies \frac{dy}{dx} = \frac{y - 4x(x^2 + y^2)}{4y(x^2 + y^2) - x} \)
\( \implies \frac{dy}{dx} = \frac{y - 4x^3 - 4xy^2}{4x^2y + 4y^3 - x} \) This demonstrates how to combine chain rule and product rule effectively.
In simple words: This equation is more complex because it has a power on one side and a product on the other. For the left side, use the chain rule first, then differentiate the inner part. For the right side, use the product rule. After differentiating, group terms with \( \frac{dy}{dx} \) and solve for it.

🎯 Exam Tip: When using the chain rule on an expression like \( (f(x))^n \), it differentiates to \( n(f(x))^{n-1} \cdot f'(x) \). Don't forget the derivative of the inside function \( f'(x) \).

Question 9. \( \sqrt{x} + \sqrt{y} = \sqrt{a} \)
Answer: Given the equation \( \sqrt{x} + \sqrt{y} = \sqrt{a} \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(\sqrt{a}) \)
\( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \)
\( \implies \frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \)
\( \implies \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \cdot 2\sqrt{y} \)
\( \implies \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \)
\( \implies \frac{dy}{dx} = -\sqrt{\frac{y}{x}} \) This type of square root function is common in problems involving distances.
In simple words: Remember that \( \sqrt{x} \) is the same as \( x^{1/2} \). When you differentiate, it becomes \( \frac{1}{2} x^{-1/2} \), or \( \frac{1}{2\sqrt{x}} \). Do this for both \( x \) and \( y \), remembering to add \( \frac{dy}{dx} \) for the \( y \) term. Then, solve the equation for \( \frac{dy}{dx} \).

🎯 Exam Tip: Convert square roots to fractional exponents \( (\cdot)^{1/2} \) before differentiating, as it makes the power rule easier to apply.

Question 10. \( x^2 + y^2 = \log(xy) \)
Answer: Given the equation \( x^2 + y^2 = \log(xy) \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\log(xy)) \)
\( 2x + 2y \frac{dy}{dx} = \frac{1}{xy} \cdot \frac{d}{dx}(xy) \)
\( 2x + 2y \frac{dy}{dx} = \frac{1}{xy} \left(x \frac{dy}{dx} + y \cdot 1\right) \)
\( \implies 2x + 2y \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx} + \frac{1}{x} \)
\( \implies 2y \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} - 2x \)
\( \implies \left(2y - \frac{1}{y}\right) \frac{dy}{dx} = \frac{1 - 2x^2}{x} \)
\( \implies \frac{2y^2 - 1}{y} \frac{dy}{dx} = \frac{1 - 2x^2}{x} \)
\( \implies \frac{dy}{dx} = \frac{y(1 - 2x^2)}{x(2y^2 - 1)} \) The derivative of logarithmic functions requires special attention to the chain rule.
In simple words: Differentiate \( x^2 \) and \( y^2 \) as usual. For \( \log(xy) \), the derivative is \( \frac{1}{xy} \) multiplied by the derivative of \( xy \). Use the product rule for \( xy \). Then, gather all terms with \( \frac{dy}{dx} \) on one side and solve for it.

🎯 Exam Tip: The derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Be careful with the chain rule when \( u \) is a product like \( xy \).

Question 11. \( x^n + y^n = a^n \)
Answer: Given the equation \( x^n + y^n = a^n \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(x^n) + \frac{d}{dx}(y^n) = \frac{d}{dx}(a^n) \)
\( nx^{n-1} + ny^{n-1} \frac{dy}{dx} = 0 \)
\( \implies ny^{n-1} \frac{dy}{dx} = -nx^{n-1} \)
\( \implies \frac{dy}{dx} = -\frac{nx^{n-1}}{ny^{n-1}} \)
\( \implies \frac{dy}{dx} = -\frac{x^{n-1}}{y^{n-1}} \)
\( \implies \frac{dy}{dx} = -\left(\frac{x}{y}\right)^{n-1} \) This is a general power rule application for implicit functions.
In simple words: Use the power rule for \( x^n \) and \( y^n \). For \( y^n \), remember to multiply by \( \frac{dy}{dx} \) because \( y \) is a function of \( x \). The right side is a constant, so its derivative is zero. Then, solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: When \( n \) is a constant, the power rule \( \frac{d}{dx}(u^n) = nu^{n-1} \frac{du}{dx} \) applies. For \( x^n \), \( \frac{du}{dx} \) is 1, and for \( y^n \), \( \frac{du}{dx} \) is \( \frac{dy}{dx} \).

Question 12. \( x^{2/3} + y^{2/3} = a^{2/3} \)
Answer: Given the equation \( x^{2/3} + y^{2/3} = a^{2/3} \). We need to find the derivative with respect to \( x \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = \frac{d}{dx}(a^{2/3}) \)
\( \frac{2}{3}x^{(2/3)-1} + \frac{2}{3}y^{(2/3)-1} \frac{dy}{dx} = 0 \)
\( \implies \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0 \)
\( \implies \frac{2}{3}y^{-1/3} \frac{dy}{dx} = -\frac{2}{3}x^{-1/3} \)
\( \implies \frac{dy}{dx} = -\frac{2/3 x^{-1/3}}{2/3 y^{-1/3}} \)
\( \implies \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} \)
\( \implies \frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}} \)
\( \implies \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3} \) Fractional exponents often simplify neatly in these problems.
In simple words: Treat fractional powers like any other power. Apply the power rule to both \( x \) and \( y \) terms. For \( y \), remember to multiply by \( \frac{dy}{dx} \). The right side is a constant, so it becomes zero. Then, rearrange and simplify to get \( \frac{dy}{dx} \).

🎯 Exam Tip: Remember that \( u^{-n} = \frac{1}{u^n} \). This property is useful for simplifying expressions with negative exponents.

Question 13. If \( y = x \sin y \), prove that \( x \frac{dy}{dx} = \frac{y}{1 - x \cos y}. \)
Answer: Given the equation \( y = x \sin y \) (1). We need to prove \( x \frac{dy}{dx} = \frac{y}{1 - x \cos y} \).
Differentiating both sides of (1) with respect to \( x \), using the product rule on the right side:
\( \frac{d}{dx}(y) = \frac{d}{dx}(x \sin y) \)
\( \frac{dy}{dx} = x \frac{d}{dx}(\sin y) + \sin y \cdot \frac{d}{dx}(x) \)
\( \frac{dy}{dx} = x (\cos y \frac{dy}{dx}) + \sin y \cdot 1 \)
\( \implies \frac{dy}{dx} = x \cos y \frac{dy}{dx} + \sin y \)
\( \implies \frac{dy}{dx} - x \cos y \frac{dy}{dx} = \sin y \)
\( \implies \frac{dy}{dx} (1 - x \cos y) = \sin y \)
\( \implies \frac{dy}{dx} = \frac{\sin y}{1 - x \cos y} \)
Now, multiply both sides by \( x \):
\( \implies x \frac{dy}{dx} = \frac{x \sin y}{1 - x \cos y} \)
From the original equation (1), we know that \( y = x \sin y \), which means \( \sin y = \frac{y}{x} \).
Substitute \( \sin y = \frac{y}{x} \) into the expression:
\( \implies x \frac{dy}{dx} = \frac{x \left(\frac{y}{x}\right)}{1 - x \cos y} \)
\( \implies x \frac{dy}{dx} = \frac{y}{1 - x \cos y} \) Thus, the statement is proven.
In simple words: Start by differentiating both sides of the given equation. For the right side, remember to use the product rule. Gather all \( \frac{dy}{dx} \) terms and solve for \( \frac{dy}{dx} \). Finally, use the original equation to substitute and show that the result matches what you need to prove.

🎯 Exam Tip: When proving an identity, it's often helpful to isolate \( \frac{dy}{dx} \) first, then manipulate the expression using the original equation to match the target form.

Question 14. If \( ax^2 + 2hxy + by^2 = c \), verify that \( \frac { dy }{ dx } \cdot \frac { dx }{ dy } = 1. \)
Answer: Given the equation \( ax^2 + 2hxy + by^2 = c \). We need to verify that \( \frac{dy}{dx} \cdot \frac{dx}{dy} = 1 \).

First, find \( \frac{dy}{dx} \) by differentiating with respect to \( x \):
\( \frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy) + \frac{d}{dx}(by^2) = \frac{d}{dx}(c) \)
\( 2ax + 2h\left(x \frac{dy}{dx} + y \cdot 1\right) + 2by \frac{dy}{dx} = 0 \)
\( \implies 2ax + 2hx \frac{dy}{dx} + 2hy + 2by \frac{dy}{dx} = 0 \)
\( \implies (2hx + 2by) \frac{dy}{dx} = -2ax - 2hy \)
\( \implies 2(hx + by) \frac{dy}{dx} = -2(ax + hy) \)
\( \implies \frac{dy}{dx} = -\frac{ax + hy}{hx + by} \) (2)

Next, find \( \frac{dx}{dy} \) by differentiating with respect to \( y \) (treating \( x \) as a function of \( y \)):
\( \frac{d}{dy}(ax^2) + \frac{d}{dy}(2hxy) + \frac{d}{dy}(by^2) = \frac{d}{dy}(c) \)
\( 2ax \frac{dx}{dy} + 2h\left(x \cdot 1 + y \frac{dx}{dy}\right) + 2by = 0 \)
\( \implies 2ax \frac{dx}{dy} + 2hx + 2hy \frac{dx}{dy} + 2by = 0 \)
\( \implies (2ax + 2hy) \frac{dx}{dy} = -2hx - 2by \)
\( \implies 2(ax + hy) \frac{dx}{dy} = -2(hx + by) \)
\( \implies \frac{dx}{dy} = -\frac{hx + by}{ax + hy} \) (3)

Now, multiply \( \frac{dy}{dx} \) and \( \frac{dx}{dy} \):
\( \frac{dy}{dx} \cdot \frac{dx}{dy} = \left(-\frac{ax + hy}{hx + by}\right) \cdot \left(-\frac{hx + by}{ax + hy}\right) \)
\( \implies \frac{dy}{dx} \cdot \frac{dx}{dy} = 1 \) This fundamental property of derivatives highlights their reciprocal relationship.
In simple words: First, find \( \frac{dy}{dx} \) by differentiating the equation with respect to \( x \). Then, find \( \frac{dx}{dy} \) by differentiating the equation with respect to \( y \). Finally, multiply the two results. If you did it right, the answer should be 1.

🎯 Exam Tip: The property \( \frac{dy}{dx} \cdot \frac{dx}{dy} = 1 \) (or \( \frac{dy}{dx} = \frac{1}{dx/dy} \)) is a useful shortcut, but you must ensure that \( \frac{dx}{dy} \) is not zero. It's often safer to derive both separately if explicitly asked to verify.

Question 15. If \( \sin^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right) = c \), prove that \( \frac{d y}{d x}=\frac{y}{x}. \)
Answer: Given the equation \( \sin^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right) = c \). We need to prove \( \frac{dy}{dx} = \frac{y}{x} \).
From the given equation:
\( \frac{x^2-y^2}{x^2+y^2} = \sin c \)
Let \( k = \sin c \). Since \( c \) is a constant, \( k \) is also a constant.
So, \( \frac{x^2-y^2}{x^2+y^2} = k \)
We can rearrange this equation:
\( x^2 - y^2 = k(x^2 + y^2) \)
\( x^2 - y^2 = kx^2 + ky^2 \)
\( x^2 - kx^2 = ky^2 + y^2 \)
\( x^2(1 - k) = y^2(1 + k) \)
\( \frac{x^2}{y^2} = \frac{1 + k}{1 - k} \)
Let \( M = \frac{1 + k}{1 - k} \). Since \( k \) is a constant, \( M \) is also a constant.
So, \( \frac{x^2}{y^2} = M \)
\( x^2 = My^2 \)
Now, differentiate both sides with respect to \( x \):
\( \frac{d}{dx}(x^2) = \frac{d}{dx}(My^2) \)
\( 2x = M (2y \frac{dy}{dx}) \)
\( \implies x = My \frac{dy}{dx} \)
We know that \( M = \frac{x^2}{y^2} \). Substitute this back:
\( x = \frac{x^2}{y^2} \cdot y \frac{dy}{dx} \)
\( \implies x = \frac{x^2}{y} \frac{dy}{dx} \)
If \( x \neq 0 \), we can divide both sides by \( x \):
\( 1 = \frac{x}{y} \frac{dy}{dx} \)
\( \implies \frac{dy}{dx} = \frac{y}{x} \) Thus, the statement is proven. This shows how algebraic manipulation can simplify differentiation.
In simple words: First, get rid of the inverse sine by taking sine of both sides. This makes the fraction \( \frac{x^2-y^2}{x^2+y^2} \) equal to a constant. Rearrange this constant equation so it looks like \( x^2 \) equals a constant times \( y^2 \). Then, differentiate this simpler equation with respect to \( x \) and solve for \( \frac{dy}{dx} \). You will find that it equals \( \frac{y}{x} \).

🎯 Exam Tip: When an inverse trigonometric function equals a constant, the argument of the function is also a constant. Use algebraic manipulation (like Componendo and Dividendo, though not strictly necessary here) to simplify before differentiating.

Question 16. If \( y \log x = x - y \), prove that \( \frac{d y}{d x}=\frac{\log x}{(1+\log x)^2} \)
Answer: Given the equation \( y \log x = x - y \). We need to prove \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \).
First, rearrange the given equation to isolate \( y \):
\( y \log x + y = x \)
\( y (\log x + 1) = x \)
\( y = \frac{x}{1 + \log x} \)
Now, differentiate \( y \) with respect to \( x \) using the quotient rule \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \):
Let \( u = x \) and \( v = 1 + \log x \).
\( \frac{du}{dx} = 1 \)
\( \frac{dv}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(\log x) = 0 + \frac{1}{x} = \frac{1}{x} \)
So, \( \frac{dy}{dx} = \frac{(1 + \log x) \cdot 1 - x \cdot \frac{1}{x}}{(1 + \log x)^2} \)
\( \implies \frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2} \)
\( \implies \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \) This shows that sometimes solving for \(y\) explicitly simplifies the differentiation process.
In simple words: First, move all the \( y \) terms to one side of the equation and factor out \( y \). This lets you express \( y \) by itself. Then, use the quotient rule to differentiate this expression. Remember that the derivative of \( \log x \) is \( \frac{1}{x} \). Simplify your answer to match the proof.

🎯 Exam Tip: When an implicit equation can be easily rearranged to express \( y \) explicitly as a function of \( x \), it's often simpler to do so and then use the quotient or product rule rather than implicit differentiation.

Question 17. If \( \sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}} = 6 \), show that \( \frac{d y}{d x}=\frac{x-17 y}{17 x-y}. \)
Answer: Given the equation \( \sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}} = 6 \) (1). We need to show that \( \frac{dy}{dx} = \frac{x-17y}{17x-y} \).
Square both sides of equation (1):
\( \left(\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}\right)^2 = 6^2 \)
\( \frac{y}{x} + \frac{x}{y} + 2\sqrt{\frac{y}{x} \cdot \frac{x}{y}} = 36 \)
\( \frac{y}{x} + \frac{x}{y} + 2\sqrt{1} = 36 \)
\( \frac{y}{x} + \frac{x}{y} + 2 = 36 \)
\( \implies \frac{y}{x} + \frac{x}{y} = 34 \)
Multiply the entire equation by \( xy \) to clear the denominators:
\( xy\left(\frac{y}{x}\right) + xy\left(\frac{x}{y}\right) = 34xy \)
\( y^2 + x^2 = 34xy \)
Now, differentiate both sides with respect to \( x \):
\( \frac{d}{dx}(y^2) + \frac{d}{dx}(x^2) = \frac{d}{dx}(34xy) \)
\( 2y \frac{dy}{dx} + 2x = 34\left(x \frac{dy}{dx} + y \cdot 1\right) \)
\( \implies 2y \frac{dy}{dx} + 2x = 34x \frac{dy}{dx} + 34y \)
\( \implies 2y \frac{dy}{dx} - 34x \frac{dy}{dx} = 34y - 2x \)
\( \implies (2y - 34x) \frac{dy}{dx} = 34y - 2x \)
\( \implies \frac{dy}{dx} = \frac{34y - 2x}{2y - 34x} \)
Factor out 2 from the numerator and denominator:
\( \implies \frac{dy}{dx} = \frac{2(17y - x)}{2(y - 17x)} \)
\( \implies \frac{dy}{dx} = \frac{17y - x}{y - 17x} \)
To match the required form \( \frac{x-17y}{17x-y} \), multiply the numerator and denominator by -1:
\( \implies \frac{dy}{dx} = \frac{-(x - 17y)}{-(17x - y)} \)
\( \implies \frac{dy}{dx} = \frac{x-17y}{17x-y} \) Thus, the statement is shown to be true.
In simple words: First, square both sides of the original equation. This helps remove the square roots. Simplify the equation until you get \( x^2 + y^2 = 34xy \). Then, differentiate this simpler equation. Remember to use the product rule for \( 34xy \). Group all terms with \( \frac{dy}{dx} \) and solve for it. Finally, rearrange your answer to match the form given in the question.

🎯 Exam Tip: Squaring both sides of an equation can sometimes simplify expressions involving square roots, making them easier to differentiate. Be careful with algebraic simplification after squaring.

Question 18. Find \( \frac { dy }{ dx } \) if
(i) \( x = y \log (xy) \)
(ii) \( x \log y + y \log x = 5 \)
(iii) \( \sin (x + y) = \log (x + y) \)
(iv) \( \sin^2 x + 2 \cos y + xy = 0 \)
Answer:
(i) Given \( x = y \log(xy) \). We need to find \( \frac{dy}{dx} \).
We can rewrite \( \log(xy) \) as \( \log x + \log y \):
\( x = y(\log x + \log y) \)
Now, differentiate both sides with respect to \( x \), using the product rule on the right side:
\( \frac{d}{dx}(x) = \frac{d}{dx}(y(\log x + \log y)) \)
\( 1 = y \cdot \frac{d}{dx}(\log x + \log y) + (\log x + \log y) \cdot \frac{d}{dx}(y) \)
\( 1 = y \left(\frac{1}{x} + \frac{1}{y} \frac{dy}{dx}\right) + (\log x + \log y) \frac{dy}{dx} \)
\( 1 = \frac{y}{x} + \frac{y}{y} \frac{dy}{dx} + (\log x + \log y) \frac{dy}{dx} \)
\( 1 = \frac{y}{x} + \frac{dy}{dx} + (\log x + \log y) \frac{dy}{dx} \)
\( \implies 1 - \frac{y}{x} = \frac{dy}{dx} (1 + \log x + \log y) \)
\( \implies \frac{x-y}{x} = \frac{dy}{dx} (1 + \log(xy)) \)
\( \implies \frac{dy}{dx} = \frac{x-y}{x(1 + \log(xy))} \) This problem combines logarithmic differentiation with the product rule.

(ii) Given \( x \log y + y \log x = 5 \). We need to find \( \frac{dy}{dx} \).
Differentiate both sides with respect to \( x \), using the product rule for each term on the left side:
\( \frac{d}{dx}(x \log y) + \frac{d}{dx}(y \log x) = \frac{d}{dx}(5) \)
\( \left(x \cdot \frac{1}{y} \frac{dy}{dx} + \log y \cdot 1\right) + \left(y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}\right) = 0 \)
\( \implies \frac{x}{y} \frac{dy}{dx} + \log y + \frac{y}{x} + \log x \frac{dy}{dx} = 0 \)
\( \implies \left(\frac{x}{y} + \log x\right) \frac{dy}{dx} = -\left(\log y + \frac{y}{x}\right) \)
\( \implies \left(\frac{x + y \log x}{y}\right) \frac{dy}{dx} = -\left(\frac{x \log y + y}{x}\right) \)
\( \implies \frac{dy}{dx} = -\frac{y(x \log y + y)}{x(x + y \log x)} \) This equation involves multiple applications of the product rule.

(iii) Given \( \sin(x+y) = \log(x+y) \). We need to find \( \frac{dy}{dx} \).
Differentiate both sides with respect to \( x \):
\( \frac{d}{dx}(\sin(x+y)) = \frac{d}{dx}(\log(x+y)) \)
\( \cos(x+y) \cdot \frac{d}{dx}(x+y) = \frac{1}{x+y} \cdot \frac{d}{dx}(x+y) \)
\( \implies \cos(x+y) \left(1 + \frac{dy}{dx}\right) = \frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) \)
Rearrange the terms:
\( \implies \cos(x+y) \left(1 + \frac{dy}{dx}\right) - \frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 0 \)
\( \implies \left(1 + \frac{dy}{dx}\right) \left(\cos(x+y) - \frac{1}{x+y}\right) = 0 \)
This equation holds if either \( 1 + \frac{dy}{dx} = 0 \) or \( \cos(x+y) - \frac{1}{x+y} = 0 \).
If \( \cos(x+y) - \frac{1}{x+y} \neq 0 \), then we must have \( 1 + \frac{dy}{dx} = 0 \).
\( \implies \frac{dy}{dx} = -1 \) This problem highlights conditions under which a derivative might simplify.

(iv) Given \( \sin^2 x + 2 \cos y + xy = 0 \). We need to find \( \frac{dy}{dx} \).
Differentiate both sides with respect to \( x \):
\( \frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(2 \cos y) + \frac{d}{dx}(xy) = \frac{d}{dx}(0) \)
\( 2 \sin x \cos x + 2(-\sin y \frac{dy}{dx}) + (x \frac{dy}{dx} + y \cdot 1) = 0 \)
\( \implies \sin(2x) - 2 \sin y \frac{dy}{dx} + x \frac{dy}{dx} + y = 0 \)
\( \implies \sin(2x) + y + (x - 2 \sin y) \frac{dy}{dx} = 0 \)
\( \implies (x - 2 \sin y) \frac{dy}{dx} = -(\sin(2x) + y) \)
\( \implies \frac{dy}{dx} = -\frac{\sin(2x) + y}{x - 2 \sin y} \)
\( \implies \frac{dy}{dx} = \frac{\sin(2x) + y}{2 \sin y - x} \) This example uses the chain rule for trigonometric functions.
In simple words: For each part, differentiate every term in the equation. Remember to use the product rule for terms like \( xy \) and \( y \log x \). For functions of \( y \), like \( \log y \) or \( \cos y \), apply the chain rule by multiplying by \( \frac{dy}{dx} \). Then, collect all \( \frac{dy}{dx} \) terms, factor it out, and solve the equation to find \( \frac{dy}{dx} \).

🎯 Exam Tip: Pay close attention to the rules for differentiating trigonometric and logarithmic functions, especially when applying the chain rule for implicit differentiation. Simplify your final expressions where possible.

Question 19. If \( y = \sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \text{ to } \infty}}} \), prove that \( (2y - 1) \frac { dy }{ dx } = \frac { 1 }{ x }. \)
Answer: Given the infinite series \( y = \sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \text{ to } \infty}}} \). We need to prove \( (2y - 1) \frac{dy}{dx} = \frac{1}{x} \).
Since the series goes on forever, the expression under the first square root is the same as \( y \) itself. So we can write:
\( y = \sqrt{\log x + y} \)
Square both sides to eliminate the square root:
\( y^2 = \log x + y \)
Now, differentiate both sides with respect to \( x \):
\( \frac{d}{dx}(y^2) = \frac{d}{dx}(\log x) + \frac{d}{dx}(y) \)
\( 2y \frac{dy}{dx} = \frac{1}{x} + \frac{dy}{dx} \)
Move the \( \frac{dy}{dx} \) term to the left side:
\( \implies 2y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{x} \)
Factor out \( \frac{dy}{dx} \):
\( \implies (2y - 1) \frac{dy}{dx} = \frac{1}{x} \) Thus, the statement is proven. This is a common technique for infinite nested functions.
In simple words: When you see an infinite repeating pattern like this, you can replace the repeating part with \( y \) itself. This simplifies the equation to \( y = \sqrt{\log x + y} \). Then, square both sides to remove the square root. Differentiate the new equation and solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: For infinite series like \( y = \sqrt{f(x) + \sqrt{f(x) + \dots}} \), the key step is to recognize that \( y = \sqrt{f(x) + y} \). This simplifies the problem into a standard implicit differentiation task.

Question 20. If \( y = \sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \text{ to } \infty}}} \), prove that \( (2y - 1) \frac { dy }{ dx } + \sin x = 0. \)
Answer: Given the infinite series \( y = \sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \text{ to } \infty}}} \). We need to prove \( (2y - 1) \frac{dy}{dx} + \sin x = 0 \).
Since the series repeats infinitely, we can write the equation as:
\( y = \sqrt{\cos x + y} \)
Square both sides:
\( y^2 = \cos x + y \)
Now, differentiate both sides with respect to \( x \):
\( \frac{d}{dx}(y^2) = \frac{d}{dx}(\cos x) + \frac{d}{dx}(y) \)
\( 2y \frac{dy}{dx} = -\sin x + \frac{dy}{dx} \)
Move the \( \frac{dy}{dx} \) term to the left side:
\( \implies 2y \frac{dy}{dx} - \frac{dy}{dx} = -\sin x \)
Factor out \( \frac{dy}{dx} \):
\( \implies (2y - 1) \frac{dy}{dx} = -\sin x \)
Move \( -\sin x \) to the left side to match the proof:
\( \implies (2y - 1) \frac{dy}{dx} + \sin x = 0 \) Thus, the statement is proven. This follows the same pattern as other infinite nested functions.
In simple words: Like the previous question, change the infinite series to \( y = \sqrt{\cos x + y} \). Square both sides to remove the square root. Differentiate the new equation, remembering that the derivative of \( \cos x \) is \( -\sin x \). Group the \( \frac{dy}{dx} \) terms and rearrange the equation to match what you need to prove.

🎯 Exam Tip: Always double-check the derivative of trigonometric functions. A common error is forgetting the negative sign for the derivative of \( \cos x \).