RBSE Solutions Class 12 Maths Chapter 7 Differentiation Exercise 7.3

Question 1. Find \( \frac{dy}{dx} \) of following functions:
(i) \( 2x + 3y = \sin y \)
(ii) \( x^2 + xy + y^2 = 200 \)
Answer:
(i) To differentiate the equation \( 2x + 3y = \sin y \) with respect to \(x\), we apply implicit differentiation.
First, differentiate each term: \( \frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y) \)
This gives \( 2 + 3 \frac{dy}{dx} = \cos y \frac{dy}{dx} \).
Next, rearrange the terms to isolate \( \frac{dy}{dx} \). Move all terms containing \( \frac{dy}{dx} \) to one side and constants to the other:
\( 2 = \cos y \frac{dy}{dx} - 3 \frac{dy}{dx} \)
Factor out \( \frac{dy}{dx} \):
\( 2 = (\cos y - 3) \frac{dy}{dx} \)
Finally, solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = \frac{2}{\cos y - 3} \)
(ii) To differentiate the equation \( x^2 + xy + y^2 = 200 \) with respect to \(x\), we use implicit differentiation. Apply the derivative to each term:
\( \frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(200) \)
For \( \frac{d}{dx}(xy) \), use the product rule \( (uv)' = u'v + uv' \): \( x \frac{dy}{dx} + y \cdot 1 \).
For \( \frac{d}{dx}(y^2) \), use the chain rule: \( 2y \frac{dy}{dx} \).
The derivative of a constant \(200\) is \(0\).
So, the equation becomes:
\( 2x + \left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} = 0 \)
Combine like terms and group the terms with \( \frac{dy}{dx} \):
\( (2x + y) + (x + 2y) \frac{dy}{dx} = 0 \)
Move terms without \( \frac{dy}{dx} \) to the right side:
\( (x + 2y) \frac{dy}{dx} = -(2x + y) \)
Finally, solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = -\frac{2x + y}{x + 2y} \)
In simple words: For the first part, we found how y changes with x by moving all terms with dy/dx to one side. For the second part, we did the same, remembering to use the product rule for 'xy' and the chain rule for 'y²'.

🎯 Exam Tip: Always remember to apply the chain rule when differentiating terms involving \(y\) with respect to \(x\), treating \(y\) as a function of \(x\).

Question 2.
(i) \( \sqrt{x} + \sqrt{y} = \sqrt{a} \)
(ii) \( \tan (x + y) + \tan(x - y) \)
Answer:
(i) To find \( \frac{dy}{dx} \) for \( \sqrt{x} + \sqrt{y} = \sqrt{a} \), we differentiate implicitly with respect to \(x\).
The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).
The derivative of \( \sqrt{y} \) is \( \frac{1}{2\sqrt{y}} \frac{dy}{dx} \) (using the chain rule).
The derivative of \( \sqrt{a} \) (a constant) is \(0\).
So, we get:
\( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \)
Now, isolate \( \frac{dy}{dx} \):
\( \frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \)
Multiply both sides by \( 2\sqrt{y} \):
\( \implies \frac{dy}{dx} = -\frac{2\sqrt{y}}{2\sqrt{x}} \)
Simplify the expression:
\( \implies \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \)
This can also be written as:
\( \implies \frac{dy}{dx} = -\sqrt{\frac{y}{x}} \)
(ii) We are asked to find the derivative of \( \tan (x + y) + \tan(x - y) \). Assuming this expression is equal to a constant, say \(C\), so \( \tan (x + y) + \tan(x - y) = C \). We differentiate implicitly with respect to \(x\).
For \( \tan(x+y) \), its derivative is \( \sec^2(x+y) \left(1 + \frac{dy}{dx}\right) \).
For \( \tan(x-y) \), its derivative is \( \sec^2(x-y) \left(1 - \frac{dy}{dx}\right) \).
The derivative of a constant \(C\) is \(0\).
So, the equation becomes:
\( \sec^2(x+y) \left(1 + \frac{dy}{dx}\right) + \sec^2(x-y) \left(1 - \frac{dy}{dx}\right) = 0 \)
Expand the terms:
\( \sec^2(x+y) + \sec^2(x+y) \frac{dy}{dx} + \sec^2(x-y) - \sec^2(x-y) \frac{dy}{dx} = 0 \)
Group the terms that contain \( \frac{dy}{dx} \):
\( (\sec^2(x+y) - \sec^2(x-y)) \frac{dy}{dx} = -\sec^2(x+y) - \sec^2(x-y) \)
Factor out the negative sign from the right side:
\( (\sec^2(x+y) - \sec^2(x-y)) \frac{dy}{dx} = -(\sec^2(x+y) + \sec^2(x-y)) \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = \frac{-(\sec^2(x+y) + \sec^2(x-y))}{\sec^2(x+y) - \sec^2(x-y)} \)
To make the denominator positive, we can multiply the numerator and denominator by \( -1 \):
\( \implies \frac{dy}{dx} = \frac{\sec^2(x+y) + \sec^2(x-y)}{\sec^2(x-y) - \sec^2(x+y)} \)
In simple words: For the first part, we took the derivative of each square root term and then moved things around to find dy/dx. For the second part, we used the chain rule for the tangent functions, then collected all the dy/dx terms to find the final answer.

🎯 Exam Tip: Remember that \( \frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{dx} \) and \( \frac{d}{dx}(\tan u) = \sec^2 u \frac{du}{dx} \). These chain rule applications are key.

Question 3.
(i) \( \sin x + 2 \cos^2 y + xy = 0 \)
(ii) \( x \sqrt{y} + y \sqrt{x} = 1 \)
Answer:
(i) To find \( \frac{dy}{dx} \) for \( \sin x + 2 \cos^2 y + xy = 0 \), we differentiate implicitly with respect to \(x\).
Differentiate each term:
\( \frac{d}{dx}(\sin x) + \frac{d}{dx}(2 \cos^2 y) + \frac{d}{dx}(xy) = \frac{d}{dx}(0) \)
The derivative of \( \sin x \) is \( \cos x \).
For \( 2 \cos^2 y \), use the chain rule: \( 2 \cdot 2 \cos y (-\sin y) \frac{dy}{dx} = -4 \sin y \cos y \frac{dy}{dx} \). This can be written as \( -2 \sin(2y) \frac{dy}{dx} \).
For \( xy \), use the product rule: \( x \frac{dy}{dx} + y \cdot 1 \).
The derivative of \(0\) is \(0\).
So, the equation becomes:
\( \cos x - 2 \sin(2y) \frac{dy}{dx} + x \frac{dy}{dx} + y = 0 \)
Group the terms containing \( \frac{dy}{dx} \):
\( (\cos x + y) + (x - 2 \sin(2y)) \frac{dy}{dx} = 0 \)
Move terms without \( \frac{dy}{dx} \) to the right side:
\( (x - 2 \sin(2y)) \frac{dy}{dx} = -(\cos x + y) \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = -\frac{\cos x + y}{x - 2 \sin(2y)} \)
This can also be written as:
\( \implies \frac{dy}{dx} = \frac{\cos x + y}{2 \sin(2y) - x} \)
(ii) To find \( \frac{dy}{dx} \) for \( x \sqrt{y} + y \sqrt{x} = 1 \), we differentiate implicitly with respect to \(x\). Apply the product rule to both terms on the left side.
For \( x\sqrt{y} \): \( x \cdot \frac{1}{2\sqrt{y}} \frac{dy}{dx} + \sqrt{y} \cdot 1 \)
For \( y\sqrt{x} \): \( y \cdot \frac{1}{2\sqrt{x}} + \sqrt{x} \frac{dy}{dx} \)
The derivative of \(1\) is \(0\).
So, the equation becomes:
\( \frac{x}{2\sqrt{y}} \frac{dy}{dx} + \sqrt{y} + \frac{y}{2\sqrt{x}} + \sqrt{x} \frac{dy}{dx} = 0 \)
Group the terms with \( \frac{dy}{dx} \) on one side:
\( \left(\frac{x}{2\sqrt{y}} + \sqrt{x}\right) \frac{dy}{dx} = -\left(\sqrt{y} + \frac{y}{2\sqrt{x}}\right) \)
Find a common denominator for the terms inside the parentheses on both sides:
\( \left(\frac{x + 2\sqrt{x}\sqrt{y}}{2\sqrt{y}}\right) \frac{dy}{dx} = -\left(\frac{2\sqrt{y}\sqrt{x} + y}{2\sqrt{x}}\right) \)
\( \left(\frac{x + 2\sqrt{xy}}{2\sqrt{y}}\right) \frac{dy}{dx} = -\left(\frac{2\sqrt{xy} + y}{2\sqrt{x}}\right) \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = -\frac{2\sqrt{xy} + y}{2\sqrt{x}} \cdot \frac{2\sqrt{y}}{x + 2\sqrt{xy}} \)
\( \implies \frac{dy}{dx} = -\frac{\sqrt{y}(2\sqrt{xy} + y)}{\sqrt{x}(x + 2\sqrt{xy})} \)
Factor out common terms to simplify further:
Numerator: \( \sqrt{y}(2\sqrt{x}\sqrt{y} + (\sqrt{y})^2) = y(2\sqrt{x} + \sqrt{y}) \)
Denominator: \( \sqrt{x}((\sqrt{x})^2 + 2\sqrt{x}\sqrt{y}) = x(\sqrt{x} + 2\sqrt{y}) \)
\( \implies \frac{dy}{dx} = -\frac{y(2\sqrt{x} + \sqrt{y})}{x(\sqrt{x} + 2\sqrt{y})} \)
In simple words: For the first part, we used the chain rule for cos²y and the product rule for xy, then collected all dy/dx terms. For the second part, we used the product rule twice and then combined fractions to solve for dy/dx.

🎯 Exam Tip: Pay careful attention to algebraic manipulation when simplifying the expression after differentiation, especially when combining fractions and factoring out terms.

Question 4.
(i) \( (x^2 + y^2)^2 = xy \)
(ii) \( \sin (xy) + \frac{x}{y} = x^2 - y \)
Answer:
(i) To find \( \frac{dy}{dx} \) for \( (x^2 + y^2)^2 = xy \), we differentiate implicitly with respect to \(x\).
On the left side, use the chain rule: \( \frac{d}{dx}((x^2+y^2)^2) = 2(x^2+y^2) \cdot \frac{d}{dx}(x^2+y^2) = 2(x^2+y^2)(2x + 2y \frac{dy}{dx}) \).
On the right side, use the product rule: \( \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \cdot 1 \).
So, the equation becomes:
\( 2(x^2 + y^2) (2x + 2y \frac{dy}{dx}) = x \frac{dy}{dx} + y \)
Expand the left side:
\( 4x(x^2 + y^2) + 4y(x^2 + y^2) \frac{dy}{dx} = x \frac{dy}{dx} + y \)
Group terms with \( \frac{dy}{dx} \) on one side:
\( 4y(x^2 + y^2) \frac{dy}{dx} - x \frac{dy}{dx} = y - 4x(x^2 + y^2) \)
Factor out \( \frac{dy}{dx} \):
\( (4y(x^2 + y^2) - x) \frac{dy}{dx} = y - 4x(x^2 + y^2) \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = \frac{y - 4x(x^2 + y^2)}{4y(x^2 + y^2) - x} \)
To make the leading term in the numerator positive, we can multiply the numerator and denominator by \( -1 \):
\( \implies \frac{dy}{dx} = \frac{4x(x^2 + y^2) - y}{x - 4y(x^2 + y^2)} \)
(ii) To find \( \frac{dy}{dx} \) for \( \sin (xy) + \frac{x}{y} = x^2 - y \), we differentiate implicitly with respect to \(x\).
For \( \sin(xy) \), use the chain rule: \( \cos(xy) \cdot \frac{d}{dx}(xy) = \cos(xy) (x \frac{dy}{dx} + y) \).
For \( \frac{x}{y} \), use the quotient rule: \( \frac{y \cdot 1 - x \frac{dy}{dx}}{y^2} \).
For \( x^2 - y \): \( 2x - \frac{dy}{dx} \).
So, the equation becomes:
\( \cos(xy) (x \frac{dy}{dx} + y) + \frac{y - x \frac{dy}{dx}}{y^2} = 2x - \frac{dy}{dx} \)
Expand and simplify. To clear the fractions, multiply the entire equation by \(y^2\):
\( y^2 \cos(xy) (x \frac{dy}{dx} + y) + (y - x \frac{dy}{dx}) = y^2 (2x - \frac{dy}{dx}) \)
\( xy^2 \cos(xy) \frac{dy}{dx} + y^3 \cos(xy) + y - x \frac{dy}{dx} = 2xy^2 - y^2 \frac{dy}{dx} \)
Group all terms with \( \frac{dy}{dx} \) on the left side:
\( xy^2 \cos(xy) \frac{dy}{dx} - x \frac{dy}{dx} + y^2 \frac{dy}{dx} = 2xy^2 - y^3 \cos(xy) - y \)
Factor out \( \frac{dy}{dx} \):
\( (xy^2 \cos(xy) - x + y^2) \frac{dy}{dx} = 2xy^2 - y^3 \cos(xy) - y \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = \frac{2xy^2 - y^3 \cos(xy) - y}{xy^2 \cos(xy) - x + y^2} \)
We can factor out \(y\) from the numerator:
\( \implies \frac{dy}{dx} = \frac{y(2xy - y^2 \cos(xy) - 1)}{xy^2 \cos(xy) - x + y^2} \)
In simple words: For the first part, we used the chain rule for the power and then the product rule for 'xy', grouping all dy/dx terms. For the second part, we applied the chain rule for sin(xy) and the quotient rule for x/y, then cleared fractions and gathered dy/dx terms to solve.

🎯 Exam Tip: When differentiating complex implicit functions, try to clear denominators by multiplying by a common factor to simplify the process of grouping terms.

Question 5.
(i) \( x^3 + y^3 = 3axy \)
(ii) \( x^y + y^x = ab \)
Answer:
(i) To find \( \frac{dy}{dx} \) for \( x^3 + y^3 = 3axy \), we differentiate implicitly with respect to \(x\).
Differentiate each term:
\( \frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(3axy) \)
For \( x^3 \), the derivative is \( 3x^2 \).
For \( y^3 \), use the chain rule: \( 3y^2 \frac{dy}{dx} \).
For \( 3axy \), use the product rule: \( 3a \left(x \frac{dy}{dx} + y \cdot 1\right) \).
So, the equation becomes:
\( 3x^2 + 3y^2 \frac{dy}{dx} = 3ax \frac{dy}{dx} + 3ay \)
Group terms with \( \frac{dy}{dx} \) on the left side:
\( 3y^2 \frac{dy}{dx} - 3ax \frac{dy}{dx} = 3ay - 3x^2 \)
Factor out \( 3 \) and \( \frac{dy}{dx} \):
\( 3(y^2 - ax) \frac{dy}{dx} = 3(ay - x^2) \)
Divide both sides by \( 3 \):
\( (y^2 - ax) \frac{dy}{dx} = ay - x^2 \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax} \)
(ii) To find \( \frac{dy}{dx} \) for \( x^y + y^x = ab \), we use a technique called logarithmic differentiation. We define \(u = x^y\) and \(v = y^x\). The equation becomes \(u + v = ab\).
Differentiating \(u + v = ab\) with respect to \(x\) gives \( \frac{du}{dx} + \frac{dv}{dx} = 0 \), because \(ab\) is a constant and its derivative is \(0\).

First, find \( \frac{du}{dx} \) for \( u = x^y \). Take natural logarithms on both sides:
\( \log u = y \log x \)
Differentiate implicitly with respect to \(x\):
\( \frac{1}{u} \frac{du}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \)
\( \implies \frac{du}{dx} = u \left(\frac{y}{x} + \log x \frac{dy}{dx}\right) = x^y \left(\frac{y}{x} + \log x \frac{dy}{dx}\right) \)

Next, find \( \frac{dv}{dx} \) for \( v = y^x \). Take natural logarithms on both sides:
\( \log v = x \log y \)
Differentiate implicitly with respect to \(x\):
\( \frac{1}{v} \frac{dv}{dx} = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} \)
\( \implies \frac{dv}{dx} = v \left(\log y + \frac{x}{y} \frac{dy}{dx}\right) = y^x \left(\log y + \frac{x}{y} \frac{dy}{dx}\right) \)

Now substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into \( \frac{du}{dx} + \frac{dv}{dx} = 0 \):
\( x^y \left(\frac{y}{x} + \log x \frac{dy}{dx}\right) + y^x \left(\log y + \frac{x}{y} \frac{dy}{dx}\right) = 0 \)
Expand the terms:
\( y x^{y-1} + x^y \log x \frac{dy}{dx} + y^x \log y + x y^{x-1} \frac{dy}{dx} = 0 \)
Group terms containing \( \frac{dy}{dx} \):
\( (x^y \log x + x y^{x-1}) \frac{dy}{dx} = -(y x^{y-1} + y^x \log y) \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = -\frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}} \)
In simple words: For the first part, we differentiated each term, remembering to use the product rule for 'axy' and the chain rule for 'y³', then we solved for dy/dx. For the second part, we used a special method called logarithmic differentiation for expressions with variables in the exponent, taking logs and then differentiating.

🎯 Exam Tip: For functions like \(u^{v}\) where both base and exponent are variables, logarithmic differentiation is typically required. Introduce natural logarithms before differentiating.

Question 6.
(i) \( y = x^y \)
(ii) \( x^a . y^b = (x - y)^{a+b} \)
Answer:
(i) To find \( \frac{dy}{dx} \) for \( y = x^y \), we use logarithmic differentiation.
Take the natural logarithm of both sides:
\( \log y = \log(x^y) \)
Using logarithm properties, bring the exponent down:
\( \log y = y \log x \)
Now, differentiate both sides implicitly with respect to \(x\). Use the product rule for \( y \log x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \)
Group terms containing \( \frac{dy}{dx} \) on the left side:
\( \frac{1}{y} \frac{dy}{dx} - \log x \frac{dy}{dx} = \frac{y}{x} \)
Factor out \( \frac{dy}{dx} \):
\( \left(\frac{1}{y} - \log x\right) \frac{dy}{dx} = \frac{y}{x} \)
Find a common denominator on the left side:
\( \left(\frac{1 - y \log x}{y}\right) \frac{dy}{dx} = \frac{y}{x} \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = \frac{y}{x} \cdot \frac{y}{1 - y \log x} \)
\( \implies \frac{dy}{dx} = \frac{y^2}{x(1 - y \log x)} \)
(ii) To find \( \frac{dy}{dx} \) for \( x^a y^b = (x - y)^{a+b} \), we use logarithmic differentiation.
Take the natural logarithm of both sides:
\( \log(x^a y^b) = \log((x - y)^{a+b}) \)
Apply logarithm properties \( \log(MN) = \log M + \log N \) and \( \log(M^p) = p \log M \):
\( a \log x + b \log y = (a + b) \log(x - y) \)
Now, differentiate both sides implicitly with respect to \(x\):
\( \frac{d}{dx}(a \log x) + \frac{d}{dx}(b \log y) = \frac{d}{dx}((a + b) \log(x - y)) \)
\( a \cdot \frac{1}{x} + b \cdot \frac{1}{y} \frac{dy}{dx} = (a + b) \cdot \frac{1}{x - y} \left(1 - \frac{dy}{dx}\right) \)
\( \frac{a}{x} + \frac{b}{y} \frac{dy}{dx} = \frac{a + b}{x - y} - \frac{a + b}{x - y} \frac{dy}{dx} \)
Group all terms containing \( \frac{dy}{dx} \) on the left side:
\( \frac{b}{y} \frac{dy}{dx} + \frac{a + b}{x - y} \frac{dy}{dx} = \frac{a + b}{x - y} - \frac{a}{x} \)
Factor out \( \frac{dy}{dx} \) and find a common denominator for both sides:
\( \left(\frac{b(x - y) + y(a + b)}{y(x - y)}\right) \frac{dy}{dx} = \frac{x(a + b) - a(x - y)}{x(x - y)} \)
Simplify the numerators:
Left numerator: \( bx - by + ay + by = bx + ay \)
Right numerator: \( ax + bx - ax + ay = bx + ay \)
So, the equation becomes:
\( \frac{bx + ay}{y(x - y)} \frac{dy}{dx} = \frac{bx + ay}{x(x - y)} \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = \frac{bx + ay}{x(x - y)} \cdot \frac{y(x - y)}{bx + ay} \)
Cancel the common term \( (bx + ay) \) and \( (x-y) \):
\( \implies \frac{dy}{dx} = \frac{y}{x} \)
In simple words: For both parts, we used a trick called logarithmic differentiation. We took the 'log' of both sides first. This helps to bring down powers, making the differentiation easier. After that, we just solved for dy/dx.

🎯 Exam Tip: For expressions involving products, quotients, or powers where variables appear in exponents, taking logarithms simplifies the differentiation process significantly. Remember to simplify common factors at the end.

Question 7.
(i) \( e^x + e^{x^2} +...+ e^{x^5} \)
(ii) \( \sqrt{e^{\sqrt{x}}}, x > 0 \)
Answer:
(i) To find the derivative of \( y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5} \) with respect to \(x\), we differentiate each term separately.
For each term, we use the chain rule: \( \frac{d}{dx}(e^{f(x)}) = e^{f(x)} f'(x) \).
\( \frac{d}{dx}(e^x) = e^x \cdot 1 = e^x \)
\( \frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot 2x \)
\( \frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2 \)
\( \frac{d}{dx}(e^{x^4}) = e^{x^4} \cdot \frac{d}{dx}(x^4) = e^{x^4} \cdot 4x^3 \)
\( \frac{d}{dx}(e^{x^5}) = e^{x^5} \cdot \frac{d}{dx}(x^5) = e^{x^5} \cdot 5x^4 \)
Combine these derivatives to get \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5} \)
(ii) To find the derivative of \( y = \sqrt{e^{\sqrt{x}}} \) with respect to \(x\), we can first rewrite the expression using exponent rules.
\( y = (e^{\sqrt{x}})^{1/2} = e^{\frac{1}{2}\sqrt{x}} \)
Now, differentiate \(y\) with respect to \(x\) using the chain rule. Let \( u = \frac{1}{2}\sqrt{x} \). Then \( \frac{dy}{dx} = e^u \frac{du}{dx} \).
First, find \( \frac{du}{dx} \):
\( u = \frac{1}{2}x^{1/2} \)
\( \frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{2} x^{(1/2)-1} = \frac{1}{4} x^{-1/2} = \frac{1}{4\sqrt{x}} \)
Now, substitute this back into the derivative of \(y\):
\( \implies \frac{dy}{dx} = e^{\frac{1}{2}\sqrt{x}} \cdot \frac{1}{4\sqrt{x}} \)
We can rewrite \( e^{\frac{1}{2}\sqrt{x}} \) as \( \sqrt{e^{\sqrt{x}}} \):
\( \implies \frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}} \)
In simple words: For the first part, we took the derivative of each 'e' term one by one, using the chain rule. For the second part, we first changed the square root into a power of 1/2, then used the chain rule again, step by step, to find the derivative.

🎯 Exam Tip: When differentiating exponential functions, always remember the chain rule. For \(e^{f(x)}\), the derivative is \(e^{f(x)} \cdot f'(x)\). Simplify expressions involving roots to powers before differentiating.

Question 8. Differentiate \( y = \frac{\cos x}{\log x} \)
Answer: To differentiate \( y = \frac{\cos x}{\log x} \) with respect to \(x\), we use the quotient rule, which states that for \( y = \frac{u}{v} \), \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
Here, let \( u = \cos x \) and \( v = \log x \).
First, find the derivatives of \(u\) and \(v\):
\( \frac{du}{dx} = \frac{d}{dx}(\cos x) = -\sin x \)
\( \frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x} \)
Now, substitute these into the quotient rule formula:
\( \frac{dy}{dx} = \frac{(\log x)(-\sin x) - (\cos x)(\frac{1}{x})}{(\log x)^2} \)
Simplify the numerator:
\( \frac{dy}{dx} = \frac{-\sin x \log x - \frac{\cos x}{x}}{(\log x)^2} \)
To remove the fraction in the numerator, multiply the numerator and the denominator by \(x\):
\( \frac{dy}{dx} = \frac{x(-\sin x \log x) - x\left(\frac{\cos x}{x}\right)}{x(\log x)^2} \)
\( \frac{dy}{dx} = \frac{-x \sin x \log x - \cos x}{x(\log x)^2} \)
Factor out a negative sign from the numerator:
\( \implies \frac{dy}{dx} = -\frac{x \sin x \log x + \cos x}{x(\log x)^2} \)
In simple words: We used the division rule for differentiation (quotient rule). We took the derivative of the top part, the bottom part, and then put them into the formula. Finally, we cleaned up the answer by removing the small fraction inside the main fraction.

🎯 Exam Tip: The quotient rule is essential for differentiating fractions. Pay close attention to the order of terms and the negative sign in the numerator to avoid common errors.

Question 9.
(i) \( y\sqrt{1-x^2} = \sin^{-1}x \)
(ii) \( y\sqrt{1+x} = \sqrt{1-x} \)
Answer:
(i) To find \( \frac{dy}{dx} \) for \( y\sqrt{1-x^2} = \sin^{-1}x \), we differentiate implicitly with respect to \(x\).
On the left side, use the product rule \( (uv)' = u'v + uv' \): \( y \frac{d}{dx}(\sqrt{1-x^2}) + \sqrt{1-x^2} \frac{dy}{dx} \).
The derivative of \( \sqrt{1-x^2} \) is \( \frac{1}{2\sqrt{1-x^2}}(-2x) = -\frac{x}{\sqrt{1-x^2}} \).
On the right side, the derivative of \( \sin^{-1}x \) is \( \frac{1}{\sqrt{1-x^2}} \).
So, the equation becomes:
\( y \left(-\frac{x}{\sqrt{1-x^2}}\right) + \sqrt{1-x^2} \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \)
\( -\frac{xy}{\sqrt{1-x^2}} + \sqrt{1-x^2} \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \)
Move the term with \(xy\) to the right side:
\( \sqrt{1-x^2} \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{xy}{\sqrt{1-x^2}} \)
Combine the terms on the right side since they have a common denominator:
\( \sqrt{1-x^2} \frac{dy}{dx} = \frac{1 + xy}{\sqrt{1-x^2}} \)
Multiply both sides by \( \sqrt{1-x^2} \):
\( (\sqrt{1-x^2})(\sqrt{1-x^2}) \frac{dy}{dx} = 1 + xy \)
\( (1-x^2) \frac{dy}{dx} = 1 + xy \)
Solve for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = \frac{1 + xy}{1 - x^2} \)
(ii) To find \( \frac{dy}{dx} \) for \( y\sqrt{1+x} = \sqrt{1-x} \), we can first isolate \(y\):
\( y = \frac{\sqrt{1-x}}{\sqrt{1+x}} \)
Now, differentiate \(y\) with respect to \(x\) using the quotient rule \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
Let \( u = \sqrt{1-x} \) and \( v = \sqrt{1+x} \).
First, find their derivatives:
\( \frac{du}{dx} = \frac{1}{2\sqrt{1-x}} \cdot (-1) = -\frac{1}{2\sqrt{1-x}} \)
\( \frac{dv}{dx} = \frac{1}{2\sqrt{1+x}} \cdot (1) = \frac{1}{2\sqrt{1+x}} \)
Substitute these into the quotient rule formula:
\( \frac{dy}{dx} = \frac{\sqrt{1+x} \left(-\frac{1}{2\sqrt{1-x}}\right) - \sqrt{1-x} \left(\frac{1}{2\sqrt{1+x}}\right)}{(\sqrt{1+x})^2} \)
\( \frac{dy}{dx} = \frac{-\frac{\sqrt{1+x}}{2\sqrt{1-x}} - \frac{\sqrt{1-x}}{2\sqrt{1+x}}}{1+x} \)
Find a common denominator for the terms in the numerator, which is \( 2\sqrt{1-x}\sqrt{1+x} = 2\sqrt{1-x^2} \):
\( \frac{dy}{dx} = \frac{\frac{-(\sqrt{1+x}\sqrt{1+x}) - (\sqrt{1-x}\sqrt{1-x})}{2\sqrt{1-x^2}}}{1+x} \)
\( \frac{dy}{dx} = \frac{\frac{-(1+x) - (1-x)}{2\sqrt{1-x^2}}}{1+x} \)
Simplify the numerator of the fraction:
\( \frac{dy}{dx} = \frac{\frac{-1-x-1+x}{2\sqrt{1-x^2}}}{1+x} = \frac{\frac{-2}{2\sqrt{1-x^2}}}{1+x} \)
Simplify further:
\( \implies \frac{dy}{dx} = \frac{-1}{\sqrt{1-x^2}(1+x)} \)
In simple words: For the first part, we used the product rule on the left side and the derivative of inverse sine on the right, then we solved for dy/dx. For the second part, we first rewrote y as a fraction, then used the quotient rule to differentiate.

🎯 Exam Tip: When simplifying complex fractions after applying the quotient rule, ensure all terms in the numerator are combined correctly before dividing by the denominator squared.

Question 10.
(i) \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + ...\infty}}} \)
(ii) \( y^x + x^y + x^x = a^b \)
Answer: Solution:

🎯 Exam Tip: For infinite series differentiation, identify the repeating pattern and substitute y back into the equation to simplify the differentiation process. For functions with variables in exponents, logarithmic differentiation is often the most effective method.

Question 10.
(i) \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + ...\infty}}} \)
(ii) \( yx + x^x + xx = ab \)
Answer:
(i) (Solution for this part is not available in the provided pages.)
(ii) We are given the equation \( y^x + x^y + x^x = ab \). Let's consider three parts: \( u = x^x \), \( v = x^y \), and \( w = y^x \). Then the equation becomes \( u + v + w = ab \). Now, we differentiate this equation with respect to \( x \):
\( \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0 \) (Since \( ab \) is a constant, its derivative is 0). This is our equation (i). First, let's find \( \frac{du}{dx} \) for \( u = x^x \): We take the logarithm on both sides:
\( \log u = \log (x^x) \)
\( \log u = x \log x \) Now, we differentiate with respect to \( x \):
\( \frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 \)
\( \frac{1}{u} \frac{du}{dx} = 1 + \log x \)
\( \frac{du}{dx} = u(1 + \log x) \)
\( \frac{du}{dx} = x^x (1 + \log x) \) (This is equation (ii)). Next, let's find \( \frac{dv}{dx} \) for \( v = x^y \): We take the logarithm on both sides:
\( \log v = \log (x^y) \)
\( \log v = y \log x \) Now, we differentiate with respect to \( x \):
\( \frac{1}{v} \frac{dv}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} \)
\( \frac{dv}{dx} = v \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) \)
\( \frac{dv}{dx} = x^y \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) \) (This is equation (iii)). Finally, let's find \( \frac{dw}{dx} \) for \( w = y^x \): We take the logarithm on both sides:
\( \log w = \log (y^x) \)
\( \log w = x \log y \) Now, we differentiate with respect to \( x \):
\( \frac{1}{w} \frac{dw}{dx} = x \cdot \frac{1}{y} \frac{dy}{dx} + \log y \cdot 1 \)
\( \frac{dw}{dx} = w \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) \)
\( \frac{dw}{dx} = y^x \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) \) (This is equation (iv)). Now we substitute equations (ii), (iii), and (iv) back into equation (i):
\( x^x (1 + \log x) + x^y \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) + y^x \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) = 0 \)
Let's expand the terms:
\( x^x (1 + \log x) + yx^{y-1} + x^y \log x \frac{dy}{dx} + x y^{x-1} \frac{dy}{dx} + y^x \log y = 0 \)
Now, we group the terms with \( \frac{dy}{dx} \):
\( (x^y \log x + x y^{x-1}) \frac{dy}{dx} = -x^x (1 + \log x) - yx^{y-1} - y^x \log y \)
\( (x^y \log x + x y^{x-1}) \frac{dy}{dx} = - (x^x (1 + \log x) + yx^{y-1} + y^x \log y) \)
Finally, we solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = - \frac{x^x (1 + \log x) + yx^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}} \)
Rearranging the terms in the numerator and denominator for a clearer final form:
\( \frac{dy}{dx} = - \frac{y^x \log y + y x^{y-1} + x^x (1 + \log x)}{x y^{x-1} + x^y \log x} \)
In simple words: This problem involves differentiating terms where both the base and the power are variables. We use a special method called logarithmic differentiation for each term. First, we break the equation into three separate parts. Then, for each part, we take the logarithm and differentiate it. Finally, we put all the differentiated parts back into the original equation and solve to find \( \frac{dy}{dx} \).

🎯 Exam Tip: Remember to use logarithmic differentiation when the function is in the form \( f(x)^{g(x)} \). Also, be careful when applying the product rule and chain rule during differentiation.