Question

Differentiate y=tan^-1*(a+btanx/b-atanx)

Answer 1

Y=tan^-1(a/b+tanx/1-a/btanx)
Thus,y=tan^-1{tan(tan^-1a/b)+tanx/1-{tan(tan^-1a/b)tanx}
So,y=tan^-1a/b+x
By differentiate
Y'=1 answer

Answer 2

$\therefore \frac{dy}{dx}=1$

Step-by-step explanation:

Given that,

$y= tan^{-1}(\frac{a+btanx}{b-atanx})$

$\Rightarrow y= tan^{-1}[\frac{b(\frac ab+tanx}{b(1-\frac a b tanx)}]$        [taking common b]

$\Rightarrow y= tan^{-1}[\frac{(\frac ab+tanx}{(1-\frac a b tanx)}]$

Let, $\frac a b = tan \theta$

$\Rightarrow y= tan^{-1}[\frac{tan\theta+tanx}{(1-tan\theta tanx)}]$

We know that, $tan(a+b)=\frac{tan a+tan b}{1-tan a \ tan b}$

$\Rightarrow y= tan^{-1}[tan (\theta+x)]$

$\Rightarrow y =\theta +x$

Now putting the value of θ

$\Rightarrow y = \frac ab+x$

Now differentiating with respect to x

$\therefore\frac{dy}{dx}=\frac{d}{dx}(\frac ab+x)$

$\therefore \frac{dy}{dx}=1$