Class 12-NCERT Solutions-Chapter-05-Differentiability and Continuity-Mis

Solution:

Let \( y = (3x^2 - 9x + 5)^9 \).

Using Chain Rule:

\[ \begin{aligned} \frac{dy}{dx} &= 9(3x^2 - 9x + 5)^8 \cdot \frac{d}{dx}(3x^2 - 9x + 5) \\ &= 9(3x^2 - 9x + 5)^8 \cdot (6x - 9) \\ &= 9(3x^2 - 9x + 5)^8 \cdot 3(2x - 3) \\ &= 27(2x - 3)(3x^2 - 9x + 5)^8 \end{aligned} \]

Solution:

Let \( y = \sin^3 x + \cos^6 x \).

\[ \begin{aligned} \frac{dy}{dx} &= 3\sin^2 x \cdot \frac{d}{dx}(\sin x) + 6\cos^5 x \cdot \frac{d}{dx}(\cos x) \\ &= 3\sin^2 x \cos x + 6\cos^5 x (-\sin x) \\ &= 3\sin^2 x \cos x - 6\cos^5 x \sin x \\ &= 3\sin x \cos x (\sin x - 2\cos^4 x) \end{aligned} \]

Solution:

Let \( y = (5x)^{3 \cos 2x} \). Take log on both sides:

\( \log y = 3 \cos 2x \cdot \log(5x) \).

Differentiating w.r.t \( x \):

\[ \begin{aligned} \frac{1}{y} \frac{dy}{dx} &= 3 \left[ \cos 2x \frac{d}{dx}(\log 5x) + \log(5x) \frac{d}{dx}(\cos 2x) \right] \\ &= 3 \left[ \cos 2x \cdot \frac{1}{5x} \cdot 5 + \log(5x) (-2\sin 2x) \right] \\ &= 3 \left[ \frac{\cos 2x}{x} - 2\sin 2x \log(5x) \right] \end{aligned} \]

\( \frac{dy}{dx} = (5x)^{3 \cos 2x} \cdot 3 \left[ \frac{\cos 2x}{x} - 2\sin 2x \log(5x) \right] \).

Solution:

Let \( y = \sin^{-1}(x^{3/2}) \).

Using Chain Rule:

\[ \begin{aligned} \frac{dy}{dx} &= \frac{1}{\sqrt{1 - (x^{3/2})^2}} \cdot \frac{d}{dx}(x^{3/2}) \\ &= \frac{1}{\sqrt{1 - x^3}} \cdot \frac{3}{2}x^{1/2} \\ &= \frac{3\sqrt{x}}{2\sqrt{1 - x^3}} \end{aligned} \]

Solution:

Using Quotient Rule:

Let \( u = \cos^{-1}(x/2) \) and \( v = \sqrt{2x+7} \).

\( u' = \frac{-1}{\sqrt{1 - (x/2)^2}} \cdot \frac{1}{2} = \frac{-1}{\sqrt{4-x^2}} \).

\( v' = \frac{1}{2\sqrt{2x+7}} \cdot 2 = \frac{1}{\sqrt{2x+7}} \).

\[ \frac{dy}{dx} = \frac{\sqrt{2x+7}(\frac{-1}{\sqrt{4-x^2}}) - \cos^{-1}\frac{x}{2}(\frac{1}{\sqrt{2x+7}})}{2x+7} \]

Simplifying the numerator gives the final expression.

Solution:

We use the identity \( 1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2 \).

Since \( 0 < x < \frac{\pi}{2} \), \( \cos \frac{x}{2} > \sin \frac{x}{2} \). So \( \sqrt{1-\sin x} = \cos \frac{x}{2} - \sin \frac{x}{2} \).

Numerator: \( (\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2}) = 2\cos \frac{x}{2} \).

Denominator: \( (\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2}) = 2\sin \frac{x}{2} \).

The expression becomes \( \cot^{-1}(\frac{2\cos x/2}{2\sin x/2}) = \cot^{-1}(\cot \frac{x}{2}) = \frac{x}{2} \).

\( \frac{dy}{dx} = \frac{1}{2} \).

Solution:

Let \( y = (\log x)^{\log x} \). Take logs:

\( \log y = \log x \cdot \log(\log x) \).

Differentiating:

\[ \begin{aligned} \frac{1}{y} \frac{dy}{dx} &= \log x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot \frac{1}{x} \\ &= \frac{1}{x} [1 + \log(\log x)] \end{aligned} \]

\( \frac{dy}{dx} = \frac{(\log x)^{\log x}}{x} [1 + \log(\log x)] \).

Solution:

Using Chain Rule:

\[ \begin{aligned} \frac{dy}{dx} &= -\sin(a \cos x + b \sin x) \cdot \frac{d}{dx}(a \cos x + b \sin x) \\ &= -\sin(a \cos x + b \sin x) \cdot (-a \sin x + b \cos x) \\ &= (a \sin x - b \cos x) \sin(a \cos x + b \sin x) \end{aligned} \]

Solution:

Let \( u = \sin x - \cos x \). Then \( y = u^u \).

\( \log y = u \log u \).

\( \frac{1}{y} y' = u \cdot \frac{1}{u} u' + \log u \cdot u' = u'(1 + \log u) \).

\( u' = \cos x + \sin x \).

\( \frac{dy}{dx} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [1 + \log(\sin x - \cos x)] \).

Solution:

• \( \frac{d}{dx}(x^x) = x^x(1 + \log x) \) (Log diff)
• \( \frac{d}{dx}(x^a) = a x^{a-1} \) (Power Rule)
• \( \frac{d}{dx}(a^x) = a^x \log a \) (Exponential Rule)
• \( \frac{d}{dx}(a^a) = 0 \) (Constant)

Summing them gives the result.

Solution:

Let \( y = u + v \).

For \( u = x^{x^2-3} \): \( \log u = (x^2-3)\log x \).

\( \frac{u'}{u} = (x^2-3)\frac{1}{x} + \log x(2x) \Rightarrow u' = x^{x^2-3} [\frac{x^2-3}{x} + 2x \log x] \).

For \( v = (x-3)^{x^2} \): \( \log v = x^2 \log(x-3) \).

\( \frac{v'}{v} = x^2 \frac{1}{x-3} + \log(x-3)(2x) \Rightarrow v' = (x-3)^{x^2} [\frac{x^2}{x-3} + 2x \log(x-3)] \).

Result is \( u' + v' \).

Solution:

\( \frac{dy}{dt} = 12(\sin t) \).

\( \frac{dx}{dt} = 10(1 - \cos t) \).

\[ \frac{dy}{dx} = \frac{12 \sin t}{10(1 - \cos t)} = \frac{6}{5} \frac{2\sin(t/2)\cos(t/2)}{2\sin^2(t/2)} = \frac{6}{5} \cot \frac{t}{2} \]

Solution:

Let \( x = \sin \theta \). Then \( \sqrt{1-x^2} = \cos \theta \).

\( y = \theta + \sin^{-1}(\cos \theta) = \theta + \sin^{-1}(\sin(\frac{\pi}{2}-\theta)) = \theta + \frac{\pi}{2} - \theta = \frac{\pi}{2} \).

Since \( y \) is a constant, \( \frac{dy}{dx} = 0 \).

Solution:

\( x\sqrt{1+y} = -y\sqrt{1+x} \). Squaring both sides:

\( x^2(1+y) = y^2(1+x) \Rightarrow x^2 + x^2y = y^2 + y^2x \).

\( x^2 - y^2 = xy(y-x) \Rightarrow (x-y)(x+y) = -xy(x-y) \).

Since \( x \neq y \), divide by \( x-y \): \( x+y = -xy \Rightarrow y(1+x) = -x \Rightarrow y = \frac{-x}{1+x} \).

Differentiating using Quotient Rule:

\( \frac{dy}{dx} = - \frac{(1+x)(1) - x(1)}{(1+x)^2} = -\frac{1}{(1+x)^2} \).

Solution:

Differentiating: \( 2(x-a) + 2(y-b)y_1 = 0 \Rightarrow y_1 = -\frac{x-a}{y-b} \).

Differentiating again: \( y_2 = -\frac{(y-b)(1) - (x-a)y_1}{(y-b)^2} \).

Substitute \( y_1 \): \( y_2 = -\frac{(y-b) + \frac{(x-a)^2}{y-b}}{(y-b)^2} = -\frac{(y-b)^2 + (x-a)^2}{(y-b)^3} = -\frac{c^2}{(y-b)^3} \).

Calculate \( 1 + y_1^2 = 1 + \frac{(x-a)^2}{(y-b)^2} = \frac{c^2}{(y-b)^2} \).

Numerator: \( \left( \frac{c^2}{(y-b)^2} \right)^{3/2} = \frac{c^3}{(y-b)^3} \).

Result: \( \frac{c^3/(y-b)^3}{-c^2/(y-b)^3} = -c \), which is a constant.

Solution:

\( x = \frac{\cos y}{\cos(a+y)} \). Differentiating w.r.t \( y \):

\( \frac{dx}{dy} = \frac{\cos(a+y)(-\sin y) - \cos y(-\sin(a+y))}{\cos^2(a+y)} \).

\( \frac{dx}{dy} = \frac{\sin(a+y)\cos y - \cos(a+y)\sin y}{\cos^2(a+y)} = \frac{\sin(a+y-y)}{\cos^2(a+y)} = \frac{\sin a}{\cos^2(a+y)} \).

\( \frac{dy}{dx} = \frac{1}{dx/dy} = \frac{\cos^2(a+y)}{\sin a} \).

Solution:

\( \frac{dx}{dt} = a(-\sin t + \sin t + t \cos t) = at \cos t \).

\( \frac{dy}{dt} = a(\cos t - \cos t + t \sin t) = at \sin t \).

\( \frac{dy}{dx} = \frac{at \sin t}{at \cos t} = \tan t \).

\( \frac{d^2y}{dx^2} = \frac{d}{dx}(\tan t) = \sec^2 t \cdot \frac{dt}{dx} = \sec^2 t \cdot \frac{1}{at \cos t} = \frac{\sec^3 t}{at} \).

Solution:

\( y_1 = e^{a \cos^{-1} x} \cdot \frac{-a}{\sqrt{1-x^2}} = \frac{-ay}{\sqrt{1-x^2}} \).

Square both sides: \( (1-x^2)y_1^2 = a^2 y^2 \).

Differentiating w.r.t \( x \):

\( (1-x^2) \cdot 2y_1 y_2 + y_1^2(-2x) = a^2 \cdot 2y y_1 \).

Dividing by \( 2y_1 \):

\( (1-x^2)y_2 - x y_1 = a^2 y \Rightarrow (1-x^2)y_2 - x y_1 - a^2 y = 0 \).