NCERT Solutions Class-12-Chapter-5-Differentiability and Continuity
Excercise-5.5
Note: Logarithmic differentiation is useful for functions of the form \( y = [u(x)]^{v(x)} \) or complex products/quotients.
Steps:
1. Take natural logarithm on both sides: \( \log y = v(x) \log u(x) \).
2. Differentiate implicitly with respect to \( x \): \( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [v(x) \log u(x)] \).
3. Multiply by \( y \) to isolate \( \frac{dy}{dx} \).
Q1
Differentiate the function w.r.t. \( x \): \( \cos x \cdot \cos 2x \cdot \cos 3x \)
Solution:
Let \( y = \cos x \cos 2x \cos 3x \).
Taking logarithms on both sides:
\( \log y = \log(\cos x) + \log(\cos 2x) + \log(\cos 3x) \)
Differentiating w.r.t. \( x \):
\[ \begin{aligned} \frac{1}{y} \frac{dy}{dx} &= \frac{1}{\cos x}(-\sin x) + \frac{1}{\cos 2x}(-\sin 2x \cdot 2) + \frac{1}{\cos 3x}(-\sin 3x \cdot 3) \\ \frac{dy}{dx} &= -y [\tan x + 2\tan 2x + 3\tan 3x] \\ &= -(\cos x \cos 2x \cos 3x) [\tan x + 2\tan 2x + 3\tan 3x] \end{aligned} \]
Q2
Differentiate w.r.t. \( x \): \( \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \)
Solution:
Let \( y = \left[ \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right]^{1/2} \).
Taking logs:
\( \log y = \frac{1}{2} [\log(x-1) + \log(x-2) - \log(x-3) - \log(x-4) - \log(x-5)] \).
Differentiating:
\[ \begin{aligned} \frac{1}{y} \frac{dy}{dx} &= \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] \\ \frac{dy}{dx} &= \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] \end{aligned} \]
Q3
Differentiate w.r.t. \( x \): \( (\log x)^{\cos x} \)
Solution:
Let \( y = (\log x)^{\cos x} \).
\( \log y = \cos x \cdot \log(\log x) \).
Differentiating using product rule:
\[ \begin{aligned} \frac{1}{y} \frac{dy}{dx} &= \cos x \frac{d}{dx}(\log(\log x)) + \log(\log x) \frac{d}{dx}(\cos x) \\ &= \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x} - \sin x \log(\log x) \\ \frac{dy}{dx} &= (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right] \end{aligned} \]
Q4
Differentiate w.r.t. \( x \): \( x^x - 2^{\sin x} \)
Solution:
Let \( y = u - v \), where \( u = x^x \) and \( v = 2^{\sin x} \).
For \( u = x^x \): \( \log u = x \log x \).
\( \frac{1}{u} \frac{du}{dx} = x(\frac{1}{x}) + \log x(1) = 1 + \log x \Rightarrow \frac{du}{dx} = x^x(1 + \log x) \).
For \( v = 2^{\sin x} \): \( \frac{dv}{dx} = 2^{\sin x} (\log 2) \cos x \).
Combining results:
\( \frac{dy}{dx} = x^x(1 + \log x) - 2^{\sin x} \cos x \log 2 \).
Q5
Differentiate w.r.t. \( x \): \( (x+3)^2 \cdot (x+4)^3 \cdot (x+5)^4 \)
Solution:
Let \( y = (x+3)^2 (x+4)^3 (x+5)^4 \).
\( \log y = 2\log(x+3) + 3\log(x+4) + 4\log(x+5) \).
\[ \frac{1}{y} \frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \]
\[ \frac{dy}{dx} = (x+3)^2 (x+4)^3 (x+5)^4 \left[ \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right] \]
Q6
Differentiate w.r.t. \( x \): \( \left( x + \frac{1}{x} \right)^x + x^{\left( 1 + \frac{1}{x} \right)} \)
Solution:
Let \( y = u + v \).
Part 1 (\( u \)): \( \log u = x \log(x + \frac{1}{x}) \).
\( \frac{1}{u}\frac{du}{dx} = x \cdot \frac{1}{x + 1/x}(1 - \frac{1}{x^2}) + \log(x + \frac{1}{x}) \).
\( \frac{du}{dx} = \left( x + \frac{1}{x} \right)^x \left[ \frac{x^2-1}{x^2+1} + \log\left( x + \frac{1}{x} \right) \right] \).
Part 2 (\( v \)): \( \log v = (1 + \frac{1}{x}) \log x \).
\( \frac{1}{v}\frac{dv}{dx} = (1 + \frac{1}{x})\frac{1}{x} + \log x(-\frac{1}{x^2}) = \frac{x+1}{x^2} - \frac{\log x}{x^2} \).
\( \frac{dv}{dx} = x^{(1 + 1/x)} \left[ \frac{x+1-\log x}{x^2} \right] \).
Add \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) for the final answer.
Q7
Differentiate w.r.t. \( x \): \( (\log x)^x + x^{\log x} \)
Solution:
Let \( y = u + v \).
\( u = (\log x)^x \Rightarrow \log u = x \log(\log x) \).
\( \frac{du}{dx} = (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right] \).
\( v = x^{\log x} \Rightarrow \log v = \log x \cdot \log x = (\log x)^2 \).
\( \frac{1}{v}\frac{dv}{dx} = 2 \log x \cdot \frac{1}{x} \Rightarrow \frac{dv}{dx} = x^{\log x} \frac{2\log x}{x} \).
Result: Sum of the two derivatives.
Q8
Differentiate w.r.t. \( x \): \( (\sin x)^x + \sin^{-1}\sqrt{x} \)
Solution:
\( u = (\sin x)^x \Rightarrow \log u = x \log(\sin x) \).
\( \frac{du}{dx} = (\sin x)^x [x \cot x + \log(\sin x)] \).
\( v = \sin^{-1}\sqrt{x} \). Using chain rule:
\( \frac{dv}{dx} = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x(1-x)}} \).
Total derivative is sum of parts.
Q9
Differentiate w.r.t. \( x \): \( x^{\sin x} + (\sin x)^{\cos x} \)
Solution:
Let \( y = u + v \).
\( u = x^{\sin x} \Rightarrow u' = x^{\sin x} [\frac{\sin x}{x} + \cos x \log x] \).
\( v = (\sin x)^{\cos x} \Rightarrow v' = (\sin x)^{\cos x} [\cos x \cot x - \sin x \log(\sin x)] \).
\( \frac{dy}{dx} = u' + v' \).
Q10
Differentiate w.r.t. \( x \): \( x^{x \cos x} + \frac{x^2+1}{x^2-1} \)
Solution:
\( u = x^{x \cos x} \). Log diff: \( \log u = x \cos x \log x \).
\( \frac{du}{dx} = x^{x \cos x} [\cos x \log x - x \sin x \log x + \cos x] \).
\( v = \frac{x^2+1}{x^2-1} \). Quotient rule:
\( \frac{dv}{dx} = \frac{(x^2-1)(2x) - (x^2+1)(2x)}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2} \).
Combine terms.
Q11
Differentiate w.r.t. \( x \): \( (x \cos x)^x + (x \sin x)^{1/x} \)
Solution:
Apply log differentiation to both terms separately.
\( u = (x \cos x)^x \Rightarrow u' = (x \cos x)^x [1 - x \tan x + \log(x \cos x)] \).
\( v = (x \sin x)^{1/x} \Rightarrow v' = (x \sin x)^{1/x} \left[ \frac{x(\cot x + 1/x) - \log(x \sin x)}{x^2} \right] \).
Q12
Find \( \frac{dy}{dx} \) for \( x^y + y^x = 1 \)
Solution:
Let \( u = x^y, v = y^x \). So \( u + v = 1 \Rightarrow \frac{du}{dx} + \frac{dv}{dx} = 0 \).
\( \frac{du}{dx} = x^y (\frac{y}{x} + \log x \frac{dy}{dx}) \).
\( \frac{dv}{dx} = y^x (\frac{x}{y}\frac{dy}{dx} + \log y) \).
Substitute and collect \( \frac{dy}{dx} \) terms:
\( \frac{dy}{dx} = -\frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}} \).
Q13
Find \( \frac{dy}{dx} \) for \( y^x = x^y \)
Solution:
Take log: \( x \log y = y \log x \).
Differentiate: \( 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \).
\( \frac{dy}{dx} (\frac{x}{y} - \log x) = \frac{y}{x} - \log y \).
\( \frac{dy}{dx} = \frac{y(y - x \log y)}{x(x - y \log x)} \).
Q14
Find \( \frac{dy}{dx} \) for \( (\cos x)^y = (\cos y)^x \)
Solution:
Take log: \( y \log(\cos x) = x \log(\cos y) \).
Differentiate and group terms:
\( \frac{dy}{dx} [\log(\cos x) + x \tan y] = \log(\cos y) + y \tan x \).
\( \frac{dy}{dx} = \frac{\log(\cos y) + y \tan x}{\log(\cos x) + x \tan y} \).
Q15
Find \( \frac{dy}{dx} \) for \( xy = e^{(x-y)} \)
Solution:
Take log: \( \log x + \log y = x - y \).
Differentiate: \( \frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx} \).
\( \frac{dy}{dx} (\frac{1}{y} + 1) = 1 - \frac{1}{x} \).
\( \frac{dy}{dx} = \frac{(x-1)/x}{(y+1)/y} = \frac{y(x-1)}{x(y+1)} \).
Q16
Find the derivative of \( f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8) \) and hence find \( f'(1) \).
Solution:
\( \log f(x) = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \log(1+x^8) \).
\( \frac{f'(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \).
At \( x = 1 \):
\( f(1) = (2)(2)(2)(2) = 16 \).
\( \frac{f'(1)}{16} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} = 0.5 + 1 + 2 + 4 = 7.5 \).
\( f'(1) = 16 \times 7.5 = 120 \).
Q17
Differentiate \( (x^2 - 5x + 8)(x^3 + 7x + 9) \) by (i) Product rule, (ii) Expanding polynomial, (iii) Logarithmic differentiation.
Solution:
All three methods will yield the same result:
\( 5x^4 - 20x^3 + 45x^2 - 52x + 11 \).
(iii) Log differentiation: \( y' = y \left[ \frac{2x-5}{x^2-5x+8} + \frac{3x^2+7}{x^3+7x+9} \right] \).
Q18
If \( u, v, w \) are functions of \( x \), show that \( \frac{d}{dx}(uvw) = \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx} \).
Solution:
Method 1 (Repeated Product Rule):
\( \frac{d}{dx}(u \cdot (vw)) = u' (vw) + u (vw)' = u'vw + u(v'w + vw') = u'vw + uv'w + uvw' \).
Method 2 (Logarithmic Differentiation):
Let \( y = uvw \). \( \log y = \log u + \log v + \log w \).
\( \frac{y'}{y} = \frac{u'}{u} + \frac{v'}{v} + \frac{w'}{w} \).
\( y' = uvw (\frac{u'}{u} + \frac{v'}{v} + \frac{w'}{w}) = u'vw + uv'w + uvw' \).