Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.6

Question 1. Find \( \frac{d y}{d x} \) for the following functions:
(i) \( xy = \tan(xy) \)
(ii) \( x^2 - xy + y^2 = 7 \)
(iii) \( x^3 + y^3 + 3axy = 1 \)
Answer:
(i) Given the equation \( xy = \tan(xy) \). We need to find \( \frac{dy}{dx} \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(xy) = \frac{d}{dx} \tan(xy) \)
Using the product rule on the left side and the chain rule on the right side:
\( x \frac{dy}{dx} + y \frac{d}{dx}(x) = \sec^2(xy) \frac{d}{dx}(xy) \)
\( x \frac{dy}{dx} + y(1) = \sec^2(xy) \left(x \frac{dy}{dx} + y \frac{d}{dx}(x)\right) \)
\( x \frac{dy}{dx} + y = \sec^2(xy) \left(x \frac{dy}{dx} + y\right) \)
\( x \frac{dy}{dx} + y = x \sec^2(xy) \frac{dy}{dx} + y \sec^2(xy) \)
Now, group all terms containing \( \frac{dy}{dx} \) on one side and other terms on the other side:
\( x \frac{dy}{dx} - x \sec^2(xy) \frac{dy}{dx} = y \sec^2(xy) - y \)
Factor out \( \frac{dy}{dx} \) from the left side and \( y \) from the right side:
\( \frac{dy}{dx} [x - x \sec^2(xy)] = y[\sec^2(xy) - 1] \)
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y[\sec^2(xy) - 1]}{x[1 - \sec^2(xy)]} \)
To simplify, we can factor out \( -1 \) from the denominator:
\( \frac{dy}{dx} = \frac{y[\sec^2(xy) - 1]}{-x[\sec^2(xy) - 1]} \)
Cancel out the common term \( [\sec^2(xy) - 1] \):
\( \frac{dy}{dx} = -\frac{y}{x} \)

(ii) Given the equation \( x^2 - xy + y^2 = 7 \). We need to find \( \frac{dy}{dx} \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(x^2) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(7) \)
Apply the power rule, product rule, and chain rule where needed:
\( 2x - \left[x \frac{dy}{dx} + y \frac{d}{dx}(x)\right] + 2y \frac{dy}{dx} = 0 \)
\( 2x - \left[x \frac{dy}{dx} + y \cdot 1\right] + 2y \frac{dy}{dx} = 0 \)
\( 2x - x \frac{dy}{dx} - y + 2y \frac{dy}{dx} = 0 \)
Group terms with \( \frac{dy}{dx} \) on one side and other terms on the other side:
\( -x \frac{dy}{dx} + 2y \frac{dy}{dx} = y - 2x \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} [2y - x] = y - 2x \)
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y - 2x}{2y - x} \)

(iii) Given the equation \( x^3 + y^3 + 3axy = 1 \). We need to find \( \frac{dy}{dx} \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) + \frac{d}{dx}(3axy) = \frac{d}{dx}(1) \)
Apply the power rule, chain rule, and product rule:
\( 3x^2 + 3y^2 \frac{dy}{dx} + 3a \left(x \frac{dy}{dx} + y \cdot 1\right) = 0 \)
\( 3x^2 + 3y^2 \frac{dy}{dx} + 3ax \frac{dy}{dx} + 3ay = 0 \)
Group terms with \( \frac{dy}{dx} \) on one side and other terms on the other side:
\( 3y^2 \frac{dy}{dx} + 3ax \frac{dy}{dx} = -3x^2 - 3ay \)
Factor out \( \frac{dy}{dx} \) from the left side and \( -3 \) from the right side:
\( \frac{dy}{dx} [3y^2 + 3ax] = -3(x^2 + ay) \)
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{-3(x^2 + ay)}{3(y^2 + ax)} \)
Cancel out the common factor \( 3 \):
\( \frac{dy}{dx} = -\frac{x^2 + ay}{y^2 + ax} \)
In simple words: To find \( \frac{dy}{dx} \) (which means how fast \( y \) changes when \( x \) changes), we use differentiation rules like the product rule and chain rule. The key is to treat \( y \) as a function of \( x \) and solve for \( \frac{dy}{dx} \) after differentiating the whole equation. Remember that the derivative of a constant (like 7 or 1) is always 0.

🎯 Exam Tip: Pay close attention to the product rule \((uv)' = u'v + uv'\) and the chain rule \((f(g(x)))' = f'(g(x))g'(x)\) when differentiating implicitly. Make sure to collect all \( \frac{dy}{dx} \) terms before factoring.

Question 2. If \( x \sqrt{1+y}+y \sqrt{1+x}=0 \) and \( x \ne y \), then prove that \( \frac{d y}{d x}=-\frac{1}{(x+1)^{2}} \)
Answer:
Given the equation \( x \sqrt{1+y}+y \sqrt{1+x}=0 \). We need to prove \( \frac{d y}{d x}=-\frac{1}{(x+1)^{2}} \).
First, rearrange the given equation:
\( x \sqrt{1+y}=-y \sqrt{1+x} \)
To remove the square roots, square both sides of the equation:
\( (x \sqrt{1+y})^2 = (-y \sqrt{1+x})^2 \)
\( x^2 (1 + y) = y^2 (1 + x) \)
Expand both sides:
\( x^2 + x^2y = y^2 + y^2x \)
Move all terms to one side to form a quadratic-like expression:
\( x^2 - y^2 + x^2y - y^2x = 0 \)
Factor \( x^2 - y^2 \) as a difference of squares \( (x-y)(x+y) \). Also factor \( xy \) from the remaining terms:
\( (x + y)(x - y) + xy(x - y) = 0 \)
Factor out the common term \( (x-y) \):
\( (x-y) [(x + y) + xy] = 0 \)
This equation implies either \( x - y = 0 \) or \( x + y + xy = 0 \).
So, \( x = y \) or \( x + y + xy = 0 \).
The problem statement clearly mentions that \( x \ne y \). Therefore, we must use the second possibility:
\( x + y + xy = 0 \)
Now, we need to express \( y \) as a function of \( x \). Group terms with \( y \):
\( y + xy = -x \)
Factor out \( y \):
\( y(1 + x) = -x \)
Solve for \( y \):
\( y = \frac{-x}{1+x} \)
Now, differentiate \( y \) with respect to \( x \) using the quotient rule, which states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \). Here, \( u = -x \) and \( v = 1+x \).
\( \frac{du}{dx} = -1 \)
\( \frac{dv}{dx} = 1 \)
So, applying the quotient rule:
\( \frac{dy}{dx} = \frac{(1+x)(-1) - (-x)(1)}{(1+x)^2} \)
\( \frac{dy}{dx} = \frac{-1 - x + x}{(1+x)^2} \)
\( \frac{dy}{dx} = \frac{-1}{(1+x)^2} \)
Hence, it is proved.
In simple words: We start by changing the given equation to get rid of the square roots. Then, we use the fact that \( x \) is not equal to \( y \) to simplify the equation even more, so \( y \) can be written purely in terms of \( x \). Finally, we use a division rule (quotient rule) to find how \( y \) changes with \( x \), and we get the answer we needed to prove.

🎯 Exam Tip: When proving an identity for a derivative, first simplify the original equation as much as possible to express \( y \) explicitly in terms of \( x \), if feasible. This often makes the differentiation process simpler and less prone to errors than implicit differentiation.

Question 3. If \( 4x + 3y = \log(4x – 3y) \), then find \( \frac{d y}{d x} \)
Answer:
Given the equation \( 4x + 3y = \log(4x – 3y) \). We need to find \( \frac{dy}{dx} \).
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(4x + 3y) = \frac{d}{dx}(\log(4x – 3y)) \)
Using the chain rule for \( \log(u) \) where \( u = 4x - 3y \), so \( \frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx} \):
\( 4(1) + 3 \frac{dy}{dx} = \frac{1}{(4x-3y)} \frac{d}{dx}(4x-3y) \)
\( 4 + 3 \frac{dy}{dx} = \frac{1}{(4x-3y)} \left(4(1) - 3 \frac{dy}{dx}\right) \)
Multiply both sides by \( (4x-3y) \) to clear the denominator:
\( (4x-3y) \left(4 + 3 \frac{dy}{dx}\right) = 4 - 3 \frac{dy}{dx} \)
Expand the left side:
\( 4(4x-3y) + 3 \frac{dy}{dx} (4x-3y) = 4 - 3 \frac{dy}{dx} \)
\( 16x - 12y + 12x \frac{dy}{dx} - 9y \frac{dy}{dx} = 4 - 3 \frac{dy}{dx} \)
Group all terms containing \( \frac{dy}{dx} \) on the left side and other terms on the right side:
\( 12x \frac{dy}{dx} - 9y \frac{dy}{dx} + 3 \frac{dy}{dx} = 4 - 16x + 12y \)
Factor out \( \frac{dy}{dx} \) from the left side:
\( \frac{dy}{dx} [12x - 9y + 3] = 4 - 16x + 12y \)
Factor out 4 from the right side and 3 from the bracket on the left side:
\( 3 \frac{dy}{dx} [4x - 3y + 1] = 4[1 - 4x + 3y] \)
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{4[1 - 4x + 3y]}{3[1 + 4x - 3y]} \)
Notice that \( (1 - 4x + 3y) \) is \( -(4x - 3y - 1) \). So the terms in the numerator and denominator are almost opposite.
\( \frac{dy}{dx} = \frac{4[1 - (4x - 3y)]}{3[1 + (4x - 3y)]} \)
This is the final simplified expression for the derivative.
In simple words: We need to find how \( y \) changes with \( x \) for the given equation. We differentiate both sides, remembering that the derivative of \( \log(A) \) is \( \frac{1}{A} \) times the derivative of \( A \). After differentiating, we collect all the \( \frac{dy}{dx} \) terms together and then solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Implicit differentiation questions involving logarithmic functions often require careful application of the chain rule. Remember that \( \frac{d}{dx}(\log f(x)) = \frac{f'(x)}{f(x)} \). Also, be thorough in algebraic rearrangement to isolate \( \frac{dy}{dx} \).