Differentiation (Part-38) | S N De

Differentiation (Part-25) | S N De

 

$~13(i)~~\cos x^3 \cdot \sin^2(x^5)~~~\text{[NCERT]}$

Solution. 

let $~~y=\cos x^3 \cdot \sin^2(x^5)\\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}\left[\cos x^3 \cdot \sin^2(x^5)\right]\\=\left[\frac{d}{dx}(\cos x^3)\right] \cdot \sin^2(x^5)\\~~~~~~+\left[\frac{d}{dx}~\sin^2(x^5)\right]\cdot(\cos x^3)\\=-\sin x^3 \cdot \frac{d}{dx}(x^3) \cdot \sin^2(x^5)\\~~~~~~+2\sin(x^5)\frac{d}{dx}(\sin x^5) \cdot (\cos x^3)\\=-\sin x^3 \cdot (3x^2)\cdot \sin^2(x^5)\\~~~~~~+2\sin x^5\cdot (\cos x^5) \cdot \frac{d}{dx}(x^5) \cdot (\cos x^3)\\=-3x^2\sin x^3\cdot \sin^2(x^5)\\~~~~~~+2\sin x^5\cdot (\cos x^5) \cdot 5x^4 \cdot (\cos x^3)\\=-3x^2\sin x^3\cdot \sin^2(x^5)\\~~~~~~+10x^4 \sin x^5\cdot (\cos x^5)  \cdot (\cos x^3)~~\text{(ans.)}$

$~(ii)~~\sin^{-1}~\frac{2x}{1+x^2}$

Solution. 

let $~~y=\sin^{-1}~\frac{2x}{1+x^2} \\ \text{or,}~~ y=\sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)\\~~[~\text{where}~~x=\tan\theta~ \rightarrow(1)] \\ \text{or,}~~ y=\sin^{-1}(\sin2\theta)\\~~~~~[\because~\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}] \\ \text{or,}~~y=2\theta=2\tan^{-1}x~~[\text{By (1)}] \\ \therefore~\frac{dy}{dx}\\=2\cdot\frac{d}{dx}\left(\tan^{-1}x\right)\\=2 \cdot \frac{1}{1+x^2}\\=\frac{2}{1+x^2}~~\text{(ans.)}$

$~(iii)~~\sin^{-1}(e^{x^2})$

Solution. 

let $~~y=\sin^{-1}(e^{x^2}) \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[\sin^{-1}(e^{x^2})]\\=\frac{1}{\sqrt{1-(e^{x^2})^2}} \cdot \frac{d}{dx}(e^{x^2})\\=\frac{1}{\sqrt{1-e^{2x^2}}} \cdot e^{x^2} \cdot \frac{d}{dx}(x^2)\\=\frac{1}{\sqrt{1-e^{2x^2}}} \cdot e^{x^2} \cdot (2x)\\=\frac{2x~e^{x^2}}{\sqrt{1-e^{2x^2}}}~~\text{(ans.)}$

(iv) $~\cos^{-1}\frac{1-x}{1+x}$

Solution. 

Let $~~y=\cos^{-1}\frac{1-x}{1+x}\rightarrow(1)$

We put $~~x=\tan^2\theta~~$ so that we get from $~(1),$

$~~y=\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)\\ \text{or,}~~y=\cos^{-1}(\cos2\theta)~~[*]\\ \text{or,}~~y=2\theta=2\tan^{-1}(\sqrt x) \\ \therefore~\frac{dy}{dx}\\=2\cdot \frac{d}{dx}~[\tan^{-1}(\sqrt x)]\\=2 \cdot \frac{1}{1+(\sqrt x)^2} \cdot \frac{d}{dx}(\sqrt x)\\=\frac{2}{1+x} \cdot \frac 12 x^{\frac 12-1}\\=\frac{1}{\sqrt x(1+x)}~~\text{(ans.)}$

Note $[*]\rightarrow~~~ \because ~\cos 2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$

(v) $~\sin4x~\cos x$

Solution. 

Let $~~y=\sin4x~\cos x \\ \text{or,}~~ y=\frac 12(2 \sin4x~\cos x) \\ \text{or,}~~ y=\frac 12[\sin(4x+x)+\sin(4x-x)]\\ \text{or,}~~y=\frac 12[\sin 5x+\sin3x]\\ \therefore~\frac{dy}{dx}\\=\frac 12[\frac{d}{dx}(\sin 5x)+\frac{d}{dx}(\sin 3x)]\\=\frac 12[\cos 5x~\frac{d}{dx}(5x)+\cos3x~\frac{d}{dx}(3x)]\\=\frac 12[\cos5x~\cdot 5+\cos3x~\cdot 3]\\=\frac 12[5\cos5x+3\cos3x]~~\text{(ans.)}$

(vi) $~4\sin2x\sin6x$

Solution. 

let $~~y=4\sin2x\sin6x \\ \text{or,}~~ y=2(2\sin2x\sin6x) \\ \text{or,}~~ y=2[\cos(6x-2x)-\cos(6x+2x)]\\ \text{or,}~~ y=2[\cos4x-\cos8x]\\ \therefore~\frac{dy}{dx}\\=2\left[\frac{d}{dx}(\cos4x)-\frac{d}{dx}(\cos8x)\right]\\=2\left[-\sin4x \cdot \frac{d}{dx}(4x)\\~~~~-(-\sin8x)~\cdot \frac{d}{dx}(8x)\right]\\=2\left[-\sin4x~\cdot 4+\sin8x~\cdot8\right]\\=2 \cdot 4(-\sin4x+2\sin8x)\\=8(2\sin8x-\sin4x)~~\text{(ans.)}$

(vii) $~e^{ax}~\cos bx$

Solution. 

Let $~~y=e^{ax}~\cos bx \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}(e^{ax}~\cos bx)\\=e^{ax}~\frac{d}{dx}(\cos bx)\\~~~~~~~~+\cos bx~\frac{d}{dx}(e^{ax})\\=e^{ax}~(-\sin bx)~\cdot \frac{d}{dx}(bx)\\~~~~~~~~+\cos bx~ \cdot e^{ax} \cdot \frac{d}{dx}(ax)\\=e^{ax}~\cdot (-\sin bx) \cdot b+\cos bx~\cdot e^{ax} \cdot a\\=e^{ax}(a \cos bx-b\sin bx)~~\text{(ans.)}$

(viii) $~\sin x\sin2x\sin3x$

Solution. 

Let $~~y=\sin x\sin2x\sin3x \\ \text{or,}~~ y=\frac 12(2\sin3x\sin x)~\cdot\sin2x \\ \text{or,}~~ y=\frac 12\sin2x~[\cos(3x-x)-\cos(3x+x)] \\ \text{or,}~~ y=\frac 12 \sin2x[\cos2x-\cos4x]\\ \text{or,}~~ y=\frac 14~(2\sin2x\cos2x-2\cos4x \sin2x)\\ \text{or,}~~ y=\frac14[\sin(2 \cdot 2x)-\{\sin(4x+2x)\\~~~~-\sin(4x-2x)\}] \\ \text{or,}~~ y=\frac 14[\sin4x-(\sin6x-\sin2x)] \\ \text{or,}~~ y=\frac 14[\sin4x+\sin2x-\sin6x] \\ \therefore \frac{dy}{dx}\\=\frac 14\left[\frac{d}{dx}(\sin4x)+\frac{d}{dx}(\sin2x)\\~~~~-\frac{d}{dx}(\sin6x)\right]\\=\frac 14\left[\cos4x~\frac{d}{dx}(4x)+\cos2x~\frac{d}{dx}(2x)\\~~~~-\cos6x~\frac{d}{dx}(6x)\right] \\=\frac 14\left[\cos4x~\cdot 4+ \cos2x~\cdot 2-\cos6x~\cdot 6\right]\\=\frac 14 \cdot 2~(2\cos4x+\cos2x-3\cos6x)\\=\frac 12~(\cos2x+2\cos4x-3\cos6x)~~\text{(ans.)}$

(ix) $~\sin^px\cos^qx$

Solution. 

Let $~~y=\sin^px\cos^qx \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}(\sin^px\cos^qx)\\=\sin^px~\frac{d}{dx}(\cos^qx)\\~~~~~+\cos^qx ~\frac{d}{dx}(\sin^px)\\=\sin^px~(q\cos^{q-1}x)~\frac{d}{dx}(\cos x)\\~~~~~+\cos^qx~(p\sin^{p-1}x)~\frac{d}{dx}(\sin x)\\=q\sin^px\cos^{q-1}x \cdot(-\sin x)\\~~~~~+p\cos^qx\sin^{p-1}x~\cdot(\cos x)\\=-q\sin^{p+1}x\cos^{q-1}x+p\cos^{q+1}x\sin^{p-1}x\\=\sin^{p-1}x\cos^{q-1}x(p\cos^2x-q\sin^2x)~~\text{(ans.)}$

(x)$~\frac{\sin^{p}x}{\cos^qx}$

Solution. 

Hints : Same as $~(ix).$ We can get the  solution of $~(x)~$ from the solution of question $~(ix)~$ by replacing $~q~$ with $~(-q)~$ since $~~\frac{\sin^{p}x}{\cos^qx}=\sin^px\cdot \cos^{-q}x~$ .

From the solution of $~(ix)~$ we get,

$~~\frac{d}{dx}(\sin^px\cos^qx)\\=\sin^{p-1}x\cos^{q-1}x(p\cos^2x-q\sin^2x) \rightarrow(1)\\ ~~\text{so that}~\frac{d}{dx}(\sin^px\cos^{-q}x)\\=\sin^{p-1}x\cos^{-q-1}x(p\cos^2x-(-q)\sin^2x)\\~~~~~~[\text{By (1)}]\\=\frac{\sin^{p-1}x(p\cos^2x+q\sin^2x)}{\cos^{q+1}x}$ 

(xi) $~e^x\sin x~\cos^2x$

Solution. 

Let $~~y=e^x\sin x~\cos^2x \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}(e^x\sin x~\cos^2x)\\=\frac{d}{dx}(e^x)~\cdot(\sin x\cos^2x)\\~+e^x~\cdot\frac{d}{dx}(\sin x)\cdot \cos^2x\\~~~~+(e^x\sin x)~\cdot \frac{d}{dx}(\cos^2x)\\=e^x\sin x~\cos^2x+e^x~\cos x~\cdot \cos^2x\\~~~~+(e^x\sin x)~\cdot 2\cos x~\cdot \frac{d}{dx}(\cos x)\\=e^x\sin x~\cos^2x+e^x\cos^3x\\~~~~+e^x \sin x~\cdot 2\cos x\cdot (-\sin x)\\=e^x\sin x~\cos^2x+e^x\cos^3x\\~~~~-2e^x ~\sin^2x\cos x\\=e^x\cos x(\sin x\cos x+\cos^2x-2\sin^2x)\\=\frac{e^x\cos x}{2}(2\sin x\cos x+2\cos^2x\\~~~~-2\cdot 2\sin^2x)\\=\frac{e^x\cos x}{2}~ [\sin2x+1\\~~~~+\cos2x-2(1-\cos2x)]\\=\frac{e^x\cos x}{2}~(\sin2x+1\\~~~~+\cos2x-2+2\cos2x)\\=\frac{e^x\cos x}{2}~(\sin2x+3\cos2x-1)~~\text{(ans.)}$

$~(xii)~~\sqrt{x^2+1}-\log\left[\frac 1x+\sqrt{1+\frac{1}{x^2}}\right]$

Solution. 

Let $~~y=\sqrt{x^2+1}-\log\left[\frac 1x+\sqrt{1+\frac{1}{x^2}}\right] \\ \text{or,}~~ y=\sqrt{x^2+1}-\log\left[\frac 1x+\sqrt{\frac{x^2+1}{x^2}}\right] \\ \text{or,}~~ y=\sqrt{x^2+1}-\log\left[\frac 1x+\frac{\sqrt{x^2+1}}{x}\right] \\ \text{or,}~~ y=\sqrt{x^2+1}-\log\left(\frac{1+\sqrt{x^2+1}}{x}\right) \\ \text{or,}~~ y=(x^2+1)^{1/2}\\~~~-[\log(1+\sqrt{x^2+1})-\log x] \\ \therefore~\frac{dy}{dx}\\=\frac 12(x^2+1)^{\frac 12-1}~\cdot\frac{d}{dx}(x^2+1)\\~~~-\frac{d}{dx}~[\log(1+\sqrt{x^2+1})]\\~~~+\frac{d}{dx}(\log x)\\=\frac{1}{2\sqrt{x^2+1}} \cdot 2x\\~~~-\frac{1}{1+\sqrt{x^2+1}} \cdot \frac{d}{dx}(1+\sqrt{x^2+1})+\frac 1x\\=\frac{x}{\sqrt{x^2+1}}-\frac{1}{1+\sqrt{x^2+1}} \cdot [0+\frac{d}{dx}(x^2+1)^{1/2}]\\~~~+\frac 1x\\=\frac{x}{\sqrt{x^2+1}}-\frac{1}{1+\sqrt{x^2+1}} \cdot \frac 12 (x^2+1)^{\frac 12-1}\\~~~ \cdot \frac{d}{dx}(x^2+1)+\frac 1x\\=\frac{x}{\sqrt{x^2+1}}-\frac{1}{1+\sqrt{x^2+1}}\cdot \frac{1}{2\sqrt{x^2+1}} \cdot 2x+\frac 1x\\=\frac{x}{\sqrt{x^2+1}}\left(1-\frac{1}{1+\sqrt{x^2+1}}\right)+\frac 1x\\=\frac{x}{\sqrt{x^2+1}} \cdot \frac{1+\sqrt{x^2+1}-1}{1+\sqrt{x^2+1}}+\frac 1x\\=\frac{x}{\sqrt{x^2+1}} \cdot \frac{\sqrt{x^2+1}}{1+\sqrt{x^2+1}}+\frac 1x\\=\frac{x}{1+\sqrt{x^2+1}}+\frac 1x\\=\frac{x^2+1+\sqrt{x^2+1}}{x(1+\sqrt{x^2+1})}\\=\frac{\sqrt{x^2+1}(\sqrt{x^2+1}+1)}{x(1+\sqrt{x^2+1})}\\=\frac{\sqrt{x^2+1}}{x}~~\text{(ans.)}$