NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

Given \(\displaystyle y=5\cos x-3\sin x\). Differentiate twice using \(\displaystyle \frac{d}{dx}\cos x=-\sin x\) and \(\displaystyle \frac{d}{dx}\sin x=\cos x\).First derivative: \[\frac{dy}{dx}=5(-\sin x)-3(\cos x)=-5\sin x-3\cos x .\] Second derivative: \[\frac{d^{2}y}{dx^{2}}=-5\cos x-3(-\sin x)=-5\cos x+3\sin x .\] The key step is to recognise the bracket that reappears: \[\frac{d^{2}y}{dx^{2}}=-\big(5\cos x-3\sin x\big)=-y .\] Hence \[\frac{d^{2}y}{dx^{2}}+y=-y+y=0 .\]So \(\displaystyle \frac{d^{2}y}{dx^{2}}+y=0\), as required. \(\displaystyle \blacksquare\)

Given \(\displaystyle y=\cos^{-1}x\). Its domain is \(\displaystyle -1<x<1\) for differentiability, and then the principal value gives \(\displaystyle y\in(0,\pi)\) — remember this, because the sign of \(\displaystyle \sin y\) depends on it.Standard derivative: \[\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^{2}}}=-(1-x^{2})^{-1/2} .\] Differentiate by the chain rule: \[\frac{d^{2}y}{dx^{2}}=-\left[\left(-\frac{1}{2}\right)(1-x^{2})^{-3/2}\cdot(-2x)\right]=-\frac{x}{(1-x^{2})^{3/2}} .\]Now convert to \(\displaystyle y\). From \(\displaystyle y=\cos^{-1}x\) we have \(\displaystyle x=\cos y\) and \(\displaystyle 1-x^{2}=1-\cos^{2}y=\sin^{2}y\). Since \(\displaystyle y\in(0,\pi)\), \(\displaystyle \sin y>0\), so \[(1-x^{2})^{3/2}=\big(\sin^{2}y\big)^{3/2}=\sin^{3}y \quad(\text{not }-\sin^{3}y).\] Therefore \[\frac{d^{2}y}{dx^{2}}=-\frac{\cos y}{\sin^{3}y}=-\frac{\cos y}{\sin y}\cdot\frac{1}{\sin^{2}y} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-\cot y\,\mathrm{cosec}^{2}y,\qquad y\in(0,\pi)\ }\]

Given \(\displaystyle y=3\cos(\log x)+4\sin(\log x)\), \(\displaystyle x>0\). Write \(\displaystyle y_{1}=\frac{dy}{dx}\), \(\displaystyle y_{2}=\frac{d^{2}y}{dx^{2}}\).Differentiate by the chain rule (inner derivative \(\displaystyle \frac{1}{x}\)): \[y_{1}=-3\sin(\log x)\cdot\frac{1}{x}+4\cos(\log x)\cdot\frac{1}{x}=\frac{-3\sin(\log x)+4\cos(\log x)}{x} .\] The efficient move is to clear the denominator before differentiating again: \[x\,y_{1}=-3\sin(\log x)+4\cos(\log x).\] Differentiate both sides with respect to \(\displaystyle x\), using the product rule on the left: \[y_{1}+x\,y_{2}=-3\cos(\log x)\cdot\frac{1}{x}-4\sin(\log x)\cdot\frac{1}{x}=\frac{-\big(3\cos(\log x)+4\sin(\log x)\big)}{x}.\] Multiply through by \(\displaystyle x\): \[x\,y_{1}+x^{2}y_{2}=-\big(3\cos(\log x)+4\sin(\log x)\big)=-y ,\] the last equality being the original definition of \(\displaystyle y\). Hence \[x^{2}y_{2}+x\,y_{1}+y=0 . \qquad\blacksquare\]

Given \(\displaystyle y=\mathrm{A}e^{mx}+\mathrm{B}e^{nx}\), with \(\displaystyle \mathrm{A},\mathrm{B},m,n\) constants. Use \(\displaystyle \frac{d}{dx}e^{kx}=k\,e^{kx}\).\[\frac{dy}{dx}=\mathrm{A}m\,e^{mx}+\mathrm{B}n\,e^{nx},\qquad \frac{d^{2}y}{dx^{2}}=\mathrm{A}m^{2}e^{mx}+\mathrm{B}n^{2}e^{nx}.\] Substitute into the left side of the required identity and group the \(\displaystyle e^{mx}\) and \(\displaystyle e^{nx}\) terms separately: \[\frac{d^{2}y}{dx^{2}}-(m+n)\frac{dy}{dx}+mny =\mathrm{A}e^{mx}\big[m^{2}-(m+n)m+mn\big]+\mathrm{B}e^{nx}\big[n^{2}-(m+n)n+mn\big].\] Each bracket vanishes: \[m^{2}-m^{2}-mn+mn=0,\qquad n^{2}-mn-n^{2}+mn=0 .\] Therefore \[\frac{d^{2}y}{dx^{2}}-(m+n)\frac{dy}{dx}+mny=0 . \qquad\blacksquare\]

Given \(\displaystyle y=500e^{7x}+600e^{-7x}\). Use \(\displaystyle \frac{d}{dx}e^{kx}=k\,e^{kx}\), taking care with the sign of \(\displaystyle k=-7\).First derivative: \[\frac{dy}{dx}=500\cdot 7e^{7x}+600\cdot(-7)e^{-7x}=3500e^{7x}-4200e^{-7x}.\] Second derivative — the second \(\displaystyle -7\) makes the term positive again: \[\frac{d^{2}y}{dx^{2}}=3500\cdot 7e^{7x}-4200\cdot(-7)e^{-7x}=24500e^{7x}+29400e^{-7x}.\] Factor out \(\displaystyle 49\): \[\frac{d^{2}y}{dx^{2}}=49\big(500e^{7x}+600e^{-7x}\big)=49y .\]Hence \(\displaystyle \frac{d^{2}y}{dx^{2}}=49y\). \(\displaystyle \blacksquare\)

Given \(\displaystyle e^{y}(x+1)=1\). Since \(\displaystyle e^{y}>0\) always, \(\displaystyle x+1=e^{-y}>0\), so the relation lives on \(\displaystyle x>-1\).Differentiate both sides with respect to \(\displaystyle x\), using the product rule on the left and the chain rule on \(\displaystyle e^{y}\) (implicit differentiation: \(\displaystyle \frac{d}{dx}e^{y}=e^{y}\frac{dy}{dx}\)): \[e^{y}\frac{dy}{dx}(x+1)+e^{y}\cdot 1=0 .\] Factor \(\displaystyle e^{y}\), which is never zero, and divide it out: \[(x+1)\frac{dy}{dx}+1=0\ \Longrightarrow\ \frac{dy}{dx}=-\frac{1}{x+1}.\] Differentiate once more (power rule on \(\displaystyle (x+1)^{-1}\)): \[\frac{d^{2}y}{dx^{2}}=-\Big[-(x+1)^{-2}\Big]=\frac{1}{(x+1)^{2}} .\] But \(\displaystyle \left(\frac{dy}{dx}\right)^{2}=\left(-\frac{1}{x+1}\right)^{2}=\frac{1}{(x+1)^{2}}\) as well.Therefore \(\displaystyle \frac{d^{2}y}{dx^{2}}=\left(\frac{dy}{dx}\right)^{2}\). \(\displaystyle \blacksquare\)

Given \(\displaystyle y=\left(\tan^{-1}x\right)^{2}\). Write \(\displaystyle y_{1}=\frac{dy}{dx}\), \(\displaystyle y_{2}=\frac{d^{2}y}{dx^{2}}\).Chain rule (outer square, inner \(\displaystyle \tan^{-1}x\) with derivative \(\displaystyle \frac{1}{1+x^{2}}\)): \[y_{1}=2\tan^{-1}x\cdot\frac{1}{1+x^{2}} .\] Rather than differentiate this quotient directly, clear the denominator first: \[(1+x^{2})\,y_{1}=2\tan^{-1}x .\] Now differentiate both sides with respect to \(\displaystyle x\), product rule on the left: \[(1+x^{2})\,y_{2}+2x\,y_{1}=\frac{2}{1+x^{2}} .\] Finally multiply the whole equation by \(\displaystyle (1+x^{2})\) to remove the remaining fraction: \[(1+x^{2})^{2}y_{2}+2x(1+x^{2})\,y_{1}=2 .\]That is \(\displaystyle \left(x^{2}+1\right)^{2}y_{2}+2x\left(x^{2}+1\right)y_{1}=2\), as required. \(\displaystyle \blacksquare\)