NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

Let \(\displaystyle y=u+v\) with \(\displaystyle u=(x\cos x)^{x}\) and \(\displaystyle v=(x\sin x)^{\frac{1}{x}}\), on a set where \(\displaystyle x\cos x>0\) and \(\displaystyle x\sin x>0\) so both bases are positive.First term. \(\displaystyle \log u=x\log(x\cos x)=x\bigl[\log x+\log\cos x\bigr]\). Splitting the log of the product first makes the differentiation routine: \[\frac{1}{u}\frac{du}{dx}=\bigl[\log x+\log\cos x\bigr]+x\left[\frac{1}{x}-\tan x\right]=\log(x\cos x)+1-x\tan x.\] Hence \[\frac{du}{dx}=(x\cos x)^{x}\bigl[1-x\tan x+\log(x\cos x)\bigr].\]Second term. \(\displaystyle \log v=\dfrac{1}{x}\log(x\sin x)=\dfrac{1}{x}\bigl[\log x+\log\sin x\bigr]\). By the product rule, remembering \(\displaystyle \dfrac{d}{dx}\dfrac{1}{x}=-\dfrac{1}{x^{2}}\), \[\frac{1}{v}\frac{dv}{dx}=-\frac{1}{x^{2}}\log(x\sin x)+\frac{1}{x}\left[\frac{1}{x}+\cot x\right]=\frac{1}{x^{2}}\bigl[1+x\cot x-\log(x\sin x)\bigr].\] Hence \[\frac{dv}{dx}=(x\sin x)^{\frac{1}{x}}\cdot\frac{1}{x^{2}}\bigl[1+x\cot x-\log(x\sin x)\bigr].\]Therefore \[\frac{dy}{dx}=(x\cos x)^{x}\bigl[1-x\tan x+\log(x\cos x)\bigr]+(x\sin x)^{\frac{1}{x}}\left[\frac{1+x\cot x-\log(x\sin x)}{x^{2}}\right].\]

The relation \(\displaystyle x^{y}+y^{x}=1\) defines \(\displaystyle y\) implicitly. Neither term can be differentiated by the power rule, so write each as an exponential and differentiate implicitly, treating \(\displaystyle y\) as a function of \(\displaystyle x\).Put \(\displaystyle u=x^{y}\) and \(\displaystyle v=y^{x}\), so \(\displaystyle u+v=1\).For \(\displaystyle u\): \(\displaystyle \log u=y\log x\), hence \[\frac{1}{u}\frac{du}{dx}=\frac{dy}{dx}\log x+\frac{y}{x},\qquad \frac{du}{dx}=x^{y}\left(\log x\,\frac{dy}{dx}+\frac{y}{x}\right).\]For \(\displaystyle v\): \(\displaystyle \log v=x\log y\), hence \[\frac{1}{v}\frac{dv}{dx}=\log y+\frac{x}{y}\frac{dy}{dx},\qquad \frac{dv}{dx}=y^{x}\left(\log y+\frac{x}{y}\frac{dy}{dx}\right).\]Because the right side of the given equation is the constant \(\displaystyle 1\), \(\displaystyle \dfrac{du}{dx}+\dfrac{dv}{dx}=0\): \[x^{y}\log x\,\frac{dy}{dx}+y\,x^{y-1}+y^{x}\log y+x\,y^{x-1}\frac{dy}{dx}=0,\] using \(\displaystyle x^{y}\cdot\dfrac{y}{x}=y\,x^{y-1}\) and \(\displaystyle y^{x}\cdot\dfrac{x}{y}=x\,y^{x-1}\).Collecting the \(\displaystyle \dfrac{dy}{dx}\) terms, \[\frac{dy}{dx}\bigl(x^{y}\log x+x\,y^{x-1}\bigr)=-\bigl(y\,x^{y-1}+y^{x}\log y\bigr),\] so \[\frac{dy}{dx}=-\,\frac{y\,x^{y-1}+y^{x}\log y}{x^{y}\log x+x\,y^{x-1}}.\]

Given \(\displaystyle y^{x}=x^{y}\). Take logarithms of both sides first — this removes both variable exponents in one step: \[x\log y=y\log x.\]Now differentiate implicitly with respect to \(\displaystyle x\), applying the product rule on each side and remembering \(\displaystyle \dfrac{d}{dx}\log y=\dfrac{1}{y}\dfrac{dy}{dx}\): \[\log y+\frac{x}{y}\frac{dy}{dx}=\frac{dy}{dx}\log x+\frac{y}{x}.\]Collect the \(\displaystyle \dfrac{dy}{dx}\) terms on one side: \[\frac{dy}{dx}\left(\frac{x}{y}-\log x\right)=\frac{y}{x}-\log y,\] \[\frac{dy}{dx}\cdot\frac{x-y\log x}{y}=\frac{y-x\log y}{x}.\]Hence \[\frac{dy}{dx}=\frac{y\,(y-x\log y)}{x\,(x-y\log x)},\] valid where \(\displaystyle x>0,\ y>0\) and \(\displaystyle x-y\log x\neq 0\).

Given \(\displaystyle (\cos x)^{y}=(\cos y)^{x}\). Both sides have a variable base and a variable exponent, so take logarithms of both sides: \[y\log\cos x=x\log\cos y.\]Differentiate implicitly with respect to \(\displaystyle x\). On the left, the product rule together with \(\displaystyle \dfrac{d}{dx}\log\cos x=\dfrac{-\sin x}{\cos x}=-\tan x\); on the right, the product rule together with \(\displaystyle \dfrac{d}{dx}\log\cos y=-\tan y\,\dfrac{dy}{dx}\): \[\frac{dy}{dx}\log\cos x+y(-\tan x)=\log\cos y+x(-\tan y)\frac{dy}{dx}.\] The chain-rule factor \(\displaystyle \dfrac{dy}{dx}\) attached to \(\displaystyle \log\cos y\) is the step most often missed.Collecting the \(\displaystyle \dfrac{dy}{dx}\) terms, \[\frac{dy}{dx}\bigl(\log\cos x+x\tan y\bigr)=\log\cos y+y\tan x,\] so \[\frac{dy}{dx}=\frac{\log\cos y+y\tan x}{\log\cos x+x\tan y},\] wherever \(\displaystyle \cos x>0\), \(\displaystyle \cos y>0\) and the denominator is non-zero.

Given \(\displaystyle xy=e^{(x-y)}\), with \(\displaystyle x>0\) and \(\displaystyle y>0\) (the left side must be positive, since the right side is). Taking logarithms of both sides converts the product to a sum and strips the exponential: \[\log x+\log y=x-y.\]Differentiate implicitly with respect to \(\displaystyle x\): \[\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}.\]Collect the \(\displaystyle \dfrac{dy}{dx}\) terms: \[\frac{dy}{dx}\left(\frac{1}{y}+1\right)=1-\frac{1}{x},\qquad \frac{dy}{dx}\cdot\frac{1+y}{y}=\frac{x-1}{x}.\]Hence \[\frac{dy}{dx}=\frac{y\,(x-1)}{x\,(1+y)}.\]

A product of four factors, so use logarithmic differentiation.Let \(\displaystyle f(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})\). Each factor is positive for \(\displaystyle x>0\); more generally take logarithms of the modulus: \[\log|f(x)|=\log|1+x|+\log(1+x^{2})+\log(1+x^{4})+\log(1+x^{8}).\]Differentiating, each term by the chain rule, \[\frac{f'(x)}{f(x)}=\frac{1}{1+x}+\frac{2x}{1+x^{2}}+\frac{4x^{3}}{1+x^{4}}+\frac{8x^{7}}{1+x^{8}}.\]Hence \[f'(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})\left[\frac{1}{1+x}+\frac{2x}{1+x^{2}}+\frac{4x^{3}}{1+x^{4}}+\frac{8x^{7}}{1+x^{8}}\right].\]Now put \(\displaystyle x=1\). Substitute into the product and the bracket separately: \[f(1)=2\cdot 2\cdot 2\cdot 2=16,\] \[\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right]=\frac{1+2+4+8}{2}=\frac{15}{2}.\]Therefore \[f'(1)=16\times\frac{15}{2}=120.\]

Let \(\displaystyle u,v,w\) be differentiable functions of \(\displaystyle x\) and \(\displaystyle y=u\cdot v\cdot w\).First method: repeated use of the product rule. Group the three factors as \(\displaystyle y=(uv)\cdot w\) and apply the product rule \(\displaystyle (fg)'=f'g+fg'\) once, with \(\displaystyle f=uv\) and \(\displaystyle g=w\): \[\frac{dy}{dx}=\frac{d(uv)}{dx}\cdot w+(uv)\cdot\frac{dw}{dx}.\] Now apply the product rule again to \(\displaystyle \dfrac{d(uv)}{dx}=\dfrac{du}{dx}v+u\dfrac{dv}{dx}\): \[\frac{dy}{dx}=\left(\frac{du}{dx}v+u\frac{dv}{dx}\right)w+uv\frac{dw}{dx}=\frac{du}{dx}\,v\,w+u\,\frac{dv}{dx}\,w+u\,v\,\frac{dw}{dx}.\]Second method: logarithmic differentiation. Assume in addition that \(\displaystyle u,v,w\) are non-zero at the point considered, so that \(\displaystyle y\neq 0\) and the logarithms exist. Then \[\log|y|=\log|u|+\log|v|+\log|w|.\] Differentiating both sides with respect to \(\displaystyle x\), using \(\displaystyle \dfrac{d}{dx}\log|f|=\dfrac{f'}{f}\), \[\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}.\] Multiplying through by \(\displaystyle y=uvw\), each fraction cancels its own factor: \[\frac{dy}{dx}=uvw\left(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\right)=\frac{du}{dx}\,v\,w+u\,\frac{dv}{dx}\,w+u\,v\,\frac{dw}{dx}.\]Both routes give \[\frac{d}{dx}(u\cdot v\cdot w)=\frac{du}{dx}\,v\cdot w+u\cdot\frac{dv}{dx}\cdot w+u\cdot v\,\frac{dw}{dx},\] as required. Note the first proof needs no extra hypothesis, whereas the second is valid only where \(\displaystyle u,v,w\) are all non-zero; since both sides of the identity are continuous, the restriction is immaterial.