Question

Please explain with proper steps .​

Answer 1

Given,

$\longrightarrow\sf{y=\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)+\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)}$

Its derivative with respect to x is,

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)+\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right]}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)+\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right)\quad\quad\dots(1)}$

Let us differentiate each term.

Let,

$\longrightarrow\sf{u=\dfrac{4x}{1+5x^2}}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{d}{dx}\left(\dfrac{4x}{1+5x^2}\right)}$

By quotient rule,

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{4(1+5x^2)-4x(10x)}{(1+5x^2)^2}}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{4+20x^2-40x^2}{(1+5x^2)^2}}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{4-20x^2}{(1+5x^2)^2}}$

So,

$\longrightarrow\sf{\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)=\dfrac{d}{dx}\left(\tan^{-1}u\right)}$

By chain rule,

$\longrightarrow\sf{\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)=\dfrac{d}{du}\left(\tan^{-1}u\right)\cdot\dfrac{du}{dx}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)=\dfrac{1}{1+u^2}\cdot\dfrac{du}{dx}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)=\dfrac{1}{1+\left(\dfrac{4x}{1+5x^2}\right)^2}\cdot\dfrac{4-20x^2}{(1+5x^2)^2}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)=\dfrac{(1+5x^2)^2}{(1+5x^2)^2+16x^2}\cdot\dfrac{4-20x^2}{(1+5x^2)^2}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)=\dfrac{4-20x^2}{1+26x^2+25x^4}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)=\dfrac{4-20x^2}{1+x^2+25x^2+25x^4}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)=\dfrac{4-20x^2}{1+x^2+25x^2(1+x^2)}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)\right)=\dfrac{4-20x^2}{(1+x^2)(1+25x^2)}}$

And let,

$\longrightarrow\sf{v=\dfrac{2+3x}{3-2x}}$

$\longrightarrow\sf{\dfrac{dv}{dx}=\dfrac{d}{dx}\left(\dfrac{2+3x}{3-2x}\right)}$

By quotient rule,

$\longrightarrow\sf{\dfrac{dv}{dx}=\dfrac{3(3-2x)+2(2+3x)}{(3-2x)^2}}$

$\longrightarrow\sf{\dfrac{dv}{dx}=\dfrac{9-6x+4+6x}{(3-2x)^2}}$

$\longrightarrow\sf{\dfrac{dv}{dx}=\dfrac{13}{(3-2x)^2}}$

So,

$\longrightarrow\sf{\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right]=\dfrac{d}{dx}\left[\tan^{-1}v\right]}$

By chain rule,

$\longrightarrow\sf{\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right]=\dfrac{d}{dv}\left[\tan^{-1}v\right]\cdot\dfrac{dv}{dx}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right]=\dfrac{1}{1+v^2}\cdot\dfrac{13}{(3-2x)^2}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right]=\dfrac{1}{1+\left(\dfrac{2+3x}{3-2x}\right)^2}\cdot\dfrac{13}{(3-2x)^2}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right]=\dfrac{(3-2x)^2}{(3-2x)^2+(2+3x)^2}\cdot\dfrac{13}{(3-2x)^2}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right]=\dfrac{13}{9-12x+4x^2+4+12x+9x^2}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right]=\dfrac{13}{13+13x^2}}$

$\longrightarrow\sf{\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\right]=\dfrac{1}{1+x^2}}$

Therefore (1) becomes,

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{4-20x^2}{(1+x^2)(1+25x^2)}+\dfrac{1}{1+x^2}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{4-20x^2+1+25x^2}{(1+x^2)(1+25x^2)}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{5+5x^2}{(1+x^2)(1+25x^2)}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{5(1+x^2)}{(1+x^2)(1+25x^2)}}$

$\longrightarrow\underline{\underline{\sf{\dfrac{dy}{dx}=\dfrac{5}{1+25x^2}}}}$

Hence (4) is the answer.

Answer 2

Question :

If $\sf\:y=\tan{}^{-1}(\dfrac{4x}{1+5x^2})+\tan{}^{-1}(\dfrac{2+3x}{3-2x})$ then find $\sf\dfrac{dy}{dx}$

Formula's Used:

$\sf1)\tan{}^{-1}(\dfrac{x + y}{1-xy})=\tan{}^{-1}x+\tan{}^{-1}y$

$\sf1)\tan{}^{-1}(\dfrac{x-y}{1 +xy})=\tan{}^{-1}x-\tan{}^{-1}y$

Solution :

We have

$\sf\:y=\tan{}^{-1}(\dfrac{4x}{1+5x^2})+\tan{}^{-1}(\dfrac{2+3x}{3-2x})$

$\sf\:y=\tan{}^{-1}(\dfrac{5x-x}{1+5x^2})+\tan{}^{-1}(\dfrac{\frac{2}{3}+x}{1-\frac{2}{3}x})$

Then,

$\sf\:y=\tan^{-1}(5x)-\tan^{-1}x+\tan^{-1}(\frac{2}{3})+\tan^{-1}x)$

$\sf\:y=\tan^{-1}(5x)-\cancel{\tan^{-1}x}+\tan^{-1}(\frac{2}{3})+\cancel{\tan^{-1}x}$

$\sf\:y=\tan^{-1}(5x)+\tan^{-1}(\frac{2}{3})$

Now Differentiate, with respect to x

$\sf\dfrac{dy}{dx}=\dfrac{d(\tan{}^{-1}5x)}{d(5x)}\times\dfrac{d(5x)}{dx}+\dfrac{d(tan^{-1}\frac{2}{3})}{dx}$

We know that

$\dfrac{d(\tan^{-1}x)}{dx}=\dfrac{1}{1+x^2}$

Then ,

$\sf\dfrac{dy}{dx}=\dfrac{1}{1+25x^2}\times5+0$

$\sf\dfrac{dy}{dx}=\dfrac{5}{1+25x^2}$

It is the required solution !

$\rule{200}2$

${\underline{\sf{Differention\:Formula's}}}$

$1)\sf\:\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$2)\sf\:\frac{d(constant)}{dx} = 0$

$3) \sf\dfrac{d( \tan \: x)}{dx} = \sec{}^{2}\: x$

$\sf4) \dfrac{d(e {}^{x}) }{dx} = {e}^{x}$

•Chain rule :

Let y = f(t) ,t= g(u) and u = m(x) ,then

$\sf \dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$