NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

Chain rule for a power of a function: if \(\displaystyle y=u^{9}\) with \(\displaystyle u=3x^{2}-9x+5\), then \(\displaystyle \dfrac{dy}{dx}=9u^{8}\dfrac{du}{dx}\).Here \(\displaystyle \dfrac{du}{dx}=6x-9\), so \[\frac{dy}{dx}=9\left(3x^{2}-9x+5\right)^{8}(6x-9)=27\left(3x^{2}-9x+5\right)^{8}(2x-3). \]

Differentiate term by term, chain rule on each power.\[\frac{d}{dx}\left(\sin^{3}x\right)=3\sin^{2}x\cdot\frac{d}{dx}(\sin x)=3\sin^{2}x\cos x \] \[\frac{d}{dx}\left(\cos^{6}x\right)=6\cos^{5}x\cdot\frac{d}{dx}(\cos x)=-6\cos^{5}x\sin x \]Adding the two, \[\frac{dy}{dx}=3\sin^{2}x\cos x-6\sin x\cos^{5}x=3\sin x\cos x\left(\sin x-2\cos^{4}x\right). \]

The base \(\displaystyle 5x\) and the exponent \(\displaystyle 3\cos 2x\) both contain \(\displaystyle x\), so neither the power rule nor the exponential rule applies on its own \(\displaystyle -\) take logarithms first (logarithmic differentiation), for \(\displaystyle x>0\).Let \(\displaystyle y=(5x)^{3\cos 2x}\). Then \[\log y=3\cos 2x\,\log (5x). \]Differentiate both sides w.r.t. \(\displaystyle x\), product rule on the right and chain rule on \(\displaystyle \cos 2x\): \[\frac{1}{y}\frac{dy}{dx}=3(-2\sin 2x)\log (5x)+3\cos 2x\cdot\frac{1}{5x}\cdot 5=-6\sin 2x\,\log (5x)+\frac{3\cos 2x}{x}. \]Hence \[\frac{dy}{dx}=(5x)^{3\cos 2x}\left[\frac{3\cos 2x}{x}-6\sin 2x\,\log (5x)\right]. \]

Write \(\displaystyle x\sqrt{x}=x^{3/2}\), so \(\displaystyle y=\sin^{-1}\left(x^{3/2}\right)\); this is defined because \(\displaystyle 0\le x^{3/2}\le 1\) on \(\displaystyle 0\le x\le 1\).Chain rule with \(\displaystyle \dfrac{d}{du}\sin^{-1}u=\dfrac{1}{\sqrt{1-u^{2}}}\), taking \(\displaystyle u=x^{3/2}\), so that \(\displaystyle u^{2}=x^{3}\) and \(\displaystyle \dfrac{du}{dx}=\dfrac{3}{2}x^{1/2}\): \[\frac{dy}{dx}=\frac{1}{\sqrt{1-x^{3}}}\cdot\frac{3}{2}\sqrt{x}=\frac{3\sqrt{x}}{2\sqrt{1-x^{3}}}. \]This holds for \(\displaystyle 0\le x<1\); at \(\displaystyle x=1\) the denominator vanishes and the graph has a vertical tangent, so the function is not differentiable at the right endpoint.

Quotient rule with \(\displaystyle u=\cos^{-1}\dfrac{x}{2}\) and \(\displaystyle v=\sqrt{2x+7}\).\[\frac{du}{dx}=-\frac{1}{\sqrt{1-\dfrac{x^{2}}{4}}}\cdot\frac{1}{2}=-\frac{1}{\sqrt{4-x^{2}}},\qquad \frac{dv}{dx}=\frac{2}{2\sqrt{2x+7}}=\frac{1}{\sqrt{2x+7}} \]The factor \(\displaystyle \tfrac12\) in \(\displaystyle \dfrac{du}{dx}\) is the derivative of the inner function \(\displaystyle x/2\) \(\displaystyle -\) the step most often dropped; note also that \(\displaystyle \sqrt{4-\frac{4x^{2}}{4}}\) simplifies to \(\displaystyle \sqrt{4-x^{2}}\) only because \(\displaystyle 2\sqrt{1-\frac{x^{2}}{4}}=\sqrt{4-x^{2}}\).Then \[\frac{dy}{dx}=\frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}}=\frac{-\dfrac{\sqrt{2x+7}}{\sqrt{4-x^{2}}}-\dfrac{\cos^{-1}\dfrac{x}{2}}{\sqrt{2x+7}}}{2x+7}. \]Hence, for \(\displaystyle -2<x<2\), \[\frac{dy}{dx}=-\frac{1}{\sqrt{4-x^{2}}\,\sqrt{2x+7}}-\frac{\cos^{-1}\dfrac{x}{2}}{(2x+7)^{3/2}}. \]

Simplify the argument before differentiating, using the half-angle identities \[1+\sin x=\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2},\qquad 1-\sin x=\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^{2}. \]For \(\displaystyle 0<x<\frac{\pi}{2}\) we have \(\displaystyle 0<\frac{x}{2}<\frac{\pi}{4}\), so \(\displaystyle \cos\frac{x}{2}>\sin\frac{x}{2}>0\) and both square roots are taken positive: \[\sqrt{1+\sin x}=\cos\frac{x}{2}+\sin\frac{x}{2},\qquad \sqrt{1-\sin x}=\cos\frac{x}{2}-\sin\frac{x}{2}. \] This sign decision is the whole point of the given interval; writing \(\displaystyle \sqrt{A^{2}}=A\) without checking the quadrant is the usual error.Therefore \[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\frac{2\cos\dfrac{x}{2}}{2\sin\dfrac{x}{2}}=\cot\frac{x}{2}, \] and since \(\displaystyle \frac{x}{2}\in\left(0,\frac{\pi}{4}\right)\) lies inside the principal branch \(\displaystyle (0,\pi)\) of \(\displaystyle \cot^{-1}\), \[y=\cot^{-1}\left(\cot\frac{x}{2}\right)=\frac{x}{2}. \]Hence \[\frac{dy}{dx}=\frac{1}{2}. \]

Base and exponent both involve \(\displaystyle x\), so use logarithmic differentiation. For \(\displaystyle x>1\) we have \(\displaystyle \log x>0\), which is exactly what makes \(\displaystyle \log(\log x)\) defined.Let \(\displaystyle y=(\log x)^{\log x}\). Then \[\log y=\log x\cdot\log(\log x). \]Differentiate w.r.t. \(\displaystyle x\), product rule on the right and chain rule on \(\displaystyle \log(\log x)\): \[\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}\log(\log x)+\log x\cdot\frac{1}{\log x}\cdot\frac{1}{x}=\frac{1}{x}\left[\log(\log x)+1\right]. \]Hence \[\frac{dy}{dx}=\frac{(\log x)^{\log x}}{x}\left[1+\log(\log x)\right],\qquad x>1. \]

Chain rule with inner function \(\displaystyle u=a\cos x+b\sin x\), where \(\displaystyle a\) and \(\displaystyle b\) are constants.\[\frac{du}{dx}=-a\sin x+b\cos x \]\[\frac{dy}{dx}=-\sin u\cdot\frac{du}{dx}=-\sin\left(a\cos x+b\sin x\right)\left(b\cos x-a\sin x\right) \]That is, \[\frac{dy}{dx}=\left(a\sin x-b\cos x\right)\sin\left(a\cos x+b\sin x\right). \]

Put \(\displaystyle u=\sin x-\cos x=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\). On \(\displaystyle \frac{\pi}{4}<x<\frac{3\pi}{4}\) we have \(\displaystyle 0<x-\frac{\pi}{4}<\frac{\pi}{2}\), so \(\displaystyle u>0\) and \(\displaystyle \log u\) exists \(\displaystyle -\) that is why the interval is given.With \(\displaystyle y=u^{u}\), logarithmic differentiation gives \(\displaystyle \log y=u\log u\), so \[\frac{1}{y}\frac{dy}{dx}=\left(\log u+u\cdot\frac{1}{u}\right)\frac{du}{dx}=(1+\log u)\frac{du}{dx}, \] and \(\displaystyle \dfrac{du}{dx}=\cos x+\sin x\).Hence \[\frac{dy}{dx}=(\sin x-\cos x)^{(\sin x-\cos x)}\left(\cos x+\sin x\right)\left[1+\log\left(\sin x-\cos x\right)\right]. \]

Four terms of four different types, each needing its own rule.\(\displaystyle x^{x}\): base and exponent both variable, so log-differentiate. With \(\displaystyle u=x^{x}\), \(\displaystyle \log u=x\log x\), hence \(\displaystyle \dfrac{1}{u}\dfrac{du}{dx}=\log x+x\cdot\dfrac{1}{x}=1+\log x\) and \[\frac{d}{dx}\left(x^{x}\right)=x^{x}(1+\log x). \]\(\displaystyle x^{a}\): variable base, constant exponent, so the power rule, \(\displaystyle \dfrac{d}{dx}x^{a}=ax^{a-1}\).\(\displaystyle a^{x}\): constant base, variable exponent, so the exponential rule, \(\displaystyle \dfrac{d}{dx}a^{x}=a^{x}\log a\).\(\displaystyle a^{a}\): a pure constant, derivative \(\displaystyle 0\).Adding, \[\frac{dy}{dx}=x^{x}(1+\log x)+ax^{a-1}+a^{x}\log a. \]