Class 12 Maths NCERT Solutions Chapter 5 Ex 5.5 – Logarithmic Differentiation

Let y=\cos x\cos2x\cos3x. Take logs: \log y=\log\cos x+\log\cos2x+\log\cos3x.

Differentiate both sides w.r.t. x: \dfrac{1}{y}\dfrac{dy}{dx}=-\tan x-2\tan2x-3\tan3x.

Let y be the given function. Take logs: \log y=\dfrac{1}{2}\left[\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5)\right].

Differentiate: \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{2}\left[\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-3}-\dfrac{1}{x-4}-\dfrac{1}{x-5}\right].

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

Differentiate (product rule on the right): \dfrac{1}{y}\dfrac{dy}{dx}=-\sin x\log(\log x)+\cos x\cdot\dfrac{1}{\log x}\cdot\dfrac{1}{x}.

Let u=x^x. Taking logs: \log u=x\log x \Rightarrow \dfrac{u'}{u}=\log x+1 \Rightarrow u'=x^x(1+\log x).

Let v=2^{\sin x}. Taking logs: \log v=\sin x\log2 \Rightarrow \dfrac{v'}{v}=\cos x\log2 \Rightarrow v'=2^{\sin x}\cos x\log2.

Let y=(x+3)^2(x+4)^3(x+5)^4. Take logs: \log y=2\log(x+3)+3\log(x+4)+4\log(x+5).

Differentiate: \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{2}{x+3}+\dfrac{3}{x+4}+\dfrac{4}{x+5}.

Let u=\left(x+\frac{1}{x}\right)^x. Taking logs: \log u=x\log\left(x+\frac{1}{x}\right).

Differentiating: \dfrac{u'}{u}=\log\left(x+\frac{1}{x}\right)+\dfrac{x(1-1/x^2)}{x+1/x}=\log\left(x+\frac{1}{x}\right)+\dfrac{x^2-1}{x^2+1}.

Let v=x^{1+1/x}. Taking logs: \log v=\left(1+\frac1x\right)\log x, so \dfrac{v'}{v}=-\dfrac{\log x}{x^2}+\dfrac{1+1/x}{x}.

Let u=(\log x)^x. Taking logs: \log u=x\log(\log x), so \dfrac{u'}{u}=\log(\log x)+\dfrac{1}{\log x}.

Let v=x^{\log x}. Taking logs: \log v=(\log x)^2, so \dfrac{v'}{v}=\dfrac{2\log x}{x}.

Let u=(\sin x)^x. Taking logs: \log u=x\log\sin x, so \dfrac{u'}{u}=\log\sin x+x\cot x.

For v=\sin^{-1}\sqrt{x}: v'=\dfrac{1}{\sqrt{1-x}}\cdot\dfrac{1}{2\sqrt{x}}=\dfrac{1}{2\sqrt{x-x^2}}.

Let u=x^{\sin x}. Taking logs: \log u=\sin x\log x, so \dfrac{u'}{u}=\cos x\log x+\dfrac{\sin x}{x}.

Let v=(\sin x)^{\cos x}. Taking logs: \log v=\cos x\log\sin x, so \dfrac{v'}{v}=-\sin x\log\sin x+\dfrac{\cos^2x}{\sin x}.

Let u=x^{x\cos x}. Taking logs: \log u=x\cos x\log x. Differentiate the right side (product of three factors): \dfrac{u'}{u}=\cos x\log x-x\sin x\log x+\cos x.

Let v=\dfrac{x^2+1}{x^2-1}. By the quotient rule: v'=\dfrac{2x(x^2-1)-(x^2+1)(2x)}{(x^2-1)^2}=\dfrac{-4x}{(x^2-1)^2}.

Let u=(x\cos x)^x. Taking logs: \log u=x\left[\log x+\log\cos x\right], so \dfrac{u'}{u}=\log x+\log\cos x+1-x\tan x.

Let v=(x\sin x)^{1/x}. Taking logs: \log v=\dfrac{1}{x}\left[\log x+\log\sin x\right], so \dfrac{v'}{v}=\dfrac{1-\log(x\sin x)}{x^2}+\dfrac{\cot x}{x}.

Let u=x^y and v=y^x, so u+v=1 \Rightarrow u'+v'=0.

For u: \log u=y\log x \Rightarrow \dfrac{u'}{u}=y'\log x+\dfrac{y}{x} \Rightarrow u'=x^y\left(y'\log x+\dfrac{y}{x}\right).

For v: \log v=x\log y \Rightarrow \dfrac{v'}{v}=\log y+\dfrac{xy'}{y} \Rightarrow v'=y^x\left(\log y+\dfrac{xy'}{y}\right).

Add and set to zero, then collect y' terms: y'\left(x^y\log x+\dfrac{xy^x}{y}\right)=-\left(\dfrac{yx^y}{x}+y^x\log y\right).

Take logs of both sides directly: x\log y=y\log x.

Differentiate: \log y+\dfrac{x}{y}\dfrac{dy}{dx}=\dfrac{dy}{dx}\log x+\dfrac{y}{x}.

Collect the \dfrac{dy}{dx} terms: \dfrac{dy}{dx}\left(\dfrac{x}{y}-\log x\right)=\dfrac{y}{x}-\log y. Multiply throughout by xy: \dfrac{dy}{dx}\left(x^2-xy\log x\right)=y^2-xy\log y.

Take logs of both sides: y\log\cos x=x\log\cos y.

Differentiate: \dfrac{dy}{dx}\log\cos x+y(-\tan x)=\log\cos y+x(-\tan y)\dfrac{dy}{dx}.

Collect the \dfrac{dy}{dx} terms: \dfrac{dy}{dx}\left(\log\cos x+x\tan y\right)=\log\cos y+y\tan x.

Take logs of both sides: \log x+\log y=x-y.

Differentiate: \dfrac{1}{x}+\dfrac{1}{y}\dfrac{dy}{dx}=1-\dfrac{dy}{dx}.

Collect the \dfrac{dy}{dx} terms: \dfrac{dy}{dx}\left(\dfrac{1}{y}+1\right)=1-\dfrac{1}{x}, i.e. \dfrac{dy}{dx}\cdot\dfrac{1+y}{y}=\dfrac{x-1}{x}.

Take logs: \log f=\log(1+x)+\log(1+x^2)+\log(1+x^4)+\log(1+x^8).

Differentiate: \dfrac{f'(x)}{f(x)}=\dfrac{1}{1+x}+\dfrac{2x}{1+x^2}+\dfrac{4x^3}{1+x^4}+\dfrac{8x^7}{1+x^8}.

At x=1: f(1)=2\times2\times2\times2=16, and the bracket equals \dfrac12+\dfrac22+\dfrac42+\dfrac82=\dfrac12+1+2+4=\dfrac{15}{2}.

f'(1)=16\times\dfrac{15}{2}=120.

(i) Product rule: with u=x^2-5x+8, v=x^3+7x+9: f'=(2x-5)(x^3+7x+9)+(x^2-5x+8)(3x^2+7).

(ii) Expand first: multiplying out gives f(x)=x^5-5x^4+15x^3-26x^2+11x+72, so f'(x)=5x^4-20x^3+45x^2-52x+11.

(iii) Logarithmic differentiation: \log f=\log(x^2-5x+8)+\log(x^3+7x+9), so f'=f\left[\dfrac{2x-5}{x^2-5x+8}+\dfrac{3x^2+7}{x^3+7x+9}\right], which simplifies back to the same expression as method (i).

Method 1 — repeated product rule: let p=uv, so uvw=pw. Then \dfrac{d}{dx}(pw)=p'w+pw'=(u'v+uv')w+uvw'=u'vw+uv'w+uvw'.

Method 2 — logarithmic differentiation: let y=uvw (taking u,v,w \gt 0). Then \log y=\log u+\log v+\log w.

Differentiate: \dfrac{y'}{y}=\dfrac{u'}{u}+\dfrac{v'}{v}+\dfrac{w'}{w}, so y'=y\left(\dfrac{u'}{u}+\dfrac{v'}{v}+\dfrac{w'}{w}\right)=u'vw+uv'w+uvw'.