Differentiation (Part-38) | S N De

 Differentiation (Part-5) | S N De

 

$~(vii)~~y=\cos^{-1} \left(\frac{a+b\cos x}{b+a\cos x}\right)~~(b>a)$

Sol. $~~y=\cos^{-1} \left(\frac{a+b\cos x}{b+a\cos x}\right) \\ \text{or,}~~  \frac{dy}{dx}=-\frac{1}{\sqrt{1-\left(\frac{a+b\cos x}{b+a\cos x} \right)^2}} \times \frac{d}{dx}\left(\frac{a+b\cos x}{b+a\cos x}\right) \\ \therefore \frac{dy}{dx}\\=\frac{-(b+a\cos x)}{\sqrt{(b+a\cos x)^2-(a+b \cos x)^2}}\\~~~ \times  \frac{(b+a\cos x)(-b\sin x)-(a+b\cos x)(-a\sin x)}{(b+a\cos x)^2}\\=\frac{-1}{(b+a\cos x)} \\~~~\times \frac{-b^2\sin x-ab\sin x\cos x+a^2\sin x+ab \sin x\cos x}{\sqrt{b^2+2ab \cos x+a^2\cos x-a^2-2ab \cos x-b^2\cos ^2x}}\\=\frac{-1}{(b+a \cos x)} \times \frac{(a^2-b^2)\sin x}{\sqrt{b^2(1-\cos^2x)-a^2(1-\cos^2x)}}\\=\frac{(b^2-a^2)\sin x}{(b+a\cos x)\sqrt{b^2\sin^2x-a^2\sin^2x}}\\=\frac{(b^2-a^2)\sin x}{(b+a\cos x)} \times \frac{1}{\sqrt{\sin^2 x (b^2-a^2)}}\\=\frac{(b^2-a^2)\sin x}{(b+a\cos x)} \times \frac{1}{\sin x~\sqrt{b^2-a^2}}\\=\frac{\sqrt{b^2-a^2}}{b+a \cos x}~~\text{(ans.)}$

$~(viii)~~y=\tan^{-1} \left[\sqrt{\frac{a-b}{a+b}}~\tan\frac x2\right]$

Sol. $~y=\tan^{-1} \left[\sqrt{\frac{a-b}{a+b}}~\tan\frac x2\right] \\ \therefore \frac{dy}{dx}\\= \frac{1}{1+\frac{a-b}{a+b} ~~\tan^2 \frac x2} \times  \frac{d}{dx} \left(\sqrt{\frac{a-b}{a+b}}~\tan\frac x2\right)\\= \frac{1}{1+\frac{a-b}{a+b} ~~\tan^2 \frac x2}\times \frac 12 \sqrt{\frac{a-b}{a+b}}~~\sec^2 \frac x2\\=\frac 12 \cdot \frac{(a+b)}{(a+b)+(a-b)~\tan^2 \frac x2} \times  \sqrt{\frac{a-b}{a+b}}~~\sec^2\frac x2 \\=\frac 12 \cdot  \frac{(a+b)}{a(1+\tan^2 \frac x2)+b(1-\tan^2 \frac x2)} \times  \sqrt{\frac{a-b}{a+b}} \sec^2\frac x2 \\= \frac 12 \cdot \frac{(a+b)/(1+\tan^2 \frac x2)}{a+b~~\left(\frac{1-\tan^2 \frac x2}{1+\tan^2 \frac x2}\right)}\times \sqrt{\frac{a-b}{a+b}} \sec^2 \frac x2\\=\frac 12 \cdot  \frac{a+b}{\sec^2 \frac x2} \cdot  \frac{1}{a+b\cos x} \times  \sqrt{\frac{a-b}{a+b}} \sec^2 \frac x2 \\= \frac{\sqrt{(a+b)(a-b)}}{ a+b \cos x}\\= \frac{\sqrt{a^2-b^2}}{2(a+b\cos x)}~~\text{(ans.)}$

Note : $~~\frac{1-\tan^2 \frac x2}{1+\tan^2 \frac x2}\\=\frac{\cos^2(x/2)-\sin^2(x/2)}{\cos^2(x/2)+\sin^2(x/2)}\\= \frac{\cos (2 \times  x/2)}{1}\\=\cos x$

$~(ix)~~y=\sin \left[2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right]$

Sol. $~~y=\sin \left[2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right] \\ \text{or,}~~y=\sin \left[2~\tan^{-1}~\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}~\right] \\~~[~\text{where}~~x=\cos2\theta~ ] \\ \therefore y=\sin\left[2\tan^{-1}~\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}\right] \\ \text{or,}~~ y=\sin \left[2\tan^{-1}~\left(\frac{\sin\theta}{\cos\theta}\right)\right] \\ \text{or,}~~ y=\sin \left[2\tan^{-1}~(\tan \theta)\right] \\ \text{or,}~~ y= \sin \left(2\theta\right)\rightarrow(1)$

Since $~~x=\cos2\theta \Rightarrow \sin2\theta=\sqrt{1-x^2}\rightarrow(2)$

Hence, from $~(1)~$ and $~(2)~$ we get,

$~~~~y=\sqrt{1-x^2}=(1-x^2)^{1/2} \\ \therefore \frac{dy}{dx}=\frac 12\cdot(1-x^2)^{-1/2} \cdot (-2x) \\ \text{or,}~~\frac{dy}{dx}=\frac{-x}{\sqrt{1-x^2}}~~\text{(ans.)} $

$~(x)~~y=\sin^{-1} ~\frac{1}{\sqrt{1+x^2}}+\tan^{-1}~\left(\frac{\sqrt{1+x^2}-1}{x}\right)$

Sol. $~y=\sin^{-1} ~\frac{1}{\sqrt{1+x^2}}+\tan^{-1}~\left(\frac{\sqrt{1+x^2}-1}{x}\right)\\~~~~~~~~~~~~~~ \rightarrow(1)$

Let $~x =\cot\theta~~\text{so that}\\~~\sqrt{1+x^2}=\sqrt{1+\cot^2\theta}=\csc\theta \\ \therefore \frac{1}{\sqrt{1+x^2}}=\frac{1}{\csc \theta}=\sin\theta \\ \text{Again,}~~ \frac{\sqrt{1+x^2}-1}{x}\\=\frac{\csc\theta-1}{\cot \theta}\\=\frac{1-\sin\theta}{\cos \theta}\\=\frac{(\cos(\theta/2)-\sin(\theta/2))^2}{\cos^2(\theta/2)-\sin^2(\theta/2)}\\=\frac{[\cos(\theta/2)-\sin(\theta/2)][\cos(\theta/2)-\sin(\theta/2)]}{[\cos(\theta/2)+\sin(\theta/2)][\cos(\theta/2)-\sin(\theta/2)]}\\=\frac{\cos(\theta/2)-\sin(\theta/2)}{\cos(\theta/2)+\sin(\theta/2)}\\=\frac{1-\tan (\theta/2)}{1+\tan (\theta/2)}\\=\frac{\tan(\pi/4)-\tan (\theta/2)}{1+\tan(\pi/4) \tan(\theta/2)}\\=\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$

Hence, from $\,(1)\,$ we get,

$~~y=\sin^{-1}(\sin \theta)+\tan^{-1} \tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \\ \therefore y=\theta+\pi/4-\theta/2 \\ \text{or,}~~ y=\theta/2+\pi/4  \\ \text{or,}~~y =\frac 12 \cot^{-1}x+\pi/4 \\ \therefore \frac{dy}{dx}=-\frac{1}{2(1+x^2)}~~\text{(ans.)}$

Note : $~~x= \cot\theta \Rightarrow \theta =\cot^{-1}~x$

$~(xi)~~x^{\sin y}+y^{\sin x}=1$

Sol. $~~x^{\sin y}+y^{\sin x}=1 \\ \therefore u+v=1\\~~[~~\text{let}~~u=x^{\sin y},~~v=y^{\sin x}~] \\ \text{or,}~~\frac{du}{dx}+\frac{dv}{dx}=0\rightarrow(1)$ 

$~~u=x^{\sin y}\\ \therefore \log u=\log (x^{\sin y})\\ \text{or,}~~ \log u=\sin y \log x\\ \text{or,}~~ \frac 1u \frac{du}{dx}=(\cos y)p ~\log x+\frac{\sin y}{x}\\~~~[\text{where}~~p=\frac{dy}{dx}] \\ \therefore \frac{du}{dx}=x^{\sin y} \left((\cos y)p ~\log x+\frac{\sin y}{x}\right) \rightarrow(2) $

$~~v=y^{\sin x}\\ \therefore \log v=\log (y^{\sin x})\\ \text{or,}~~ \log v=\sin x \log y\\ \text{or,}~~ \frac 1v \frac{dv}{dx}=(\cos x) ~\log y+\frac{\sin x}{y}~p\\~~~[\text{where}~~p=\frac{dy}{dx}] \\ \therefore \frac{dv}{dx}=y^{\sin x} \left((\cos x) ~\log y+\frac{\sin x}{y}~p\right) \rightarrow(3) $

Hence, from $\,(1),\,(2),\,(3)\,$ we get,

$~~x^{\sin y} \left((\cos y)p ~\log x+\frac{\sin y}{x}\right)\\~~~~+y^{\sin x} \left((\cos x) ~\log y+\frac{\sin x}{y}~p\right)=0 \\ \text{or,}~~p\left(x^{\sin y} \cos y \log x+y^{\sin x} \frac{\sin x}{y}\right)\\~~~~=-(x^{\sin y} ~~\frac{\sin y}{x}+y^{\sin x} \cos x \log y) \\ \therefore \frac{dy}{dx}=-\frac{y}{x}\left(\frac{x^{\sin y}~\sin y+xy^{\sin x} \log y \cos x}{x^{\sin y}y \log x \cos y+y^{\sin x} \sin x}\right)$