Question

y=sec( tan -1 x).find dy\dx

Answer 1

y= (tan-1x) = dy/dx
tan=1x-y=dy/dx
tan=x-y=dy/dx
dy/dx=x-y+tan
dy/dx=tanx-y

Answer 2

Answer:

The required solution is $\frac{dy}{dx}=\frac{xy}{1+x^2}$.

Step-by-step explanation:

The given equation is

$y=sec(tan^{-1}x)$

According to the Chain rule, the derivative of f(g(x)) is

$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$

Differentiate with respect to x.

$\frac{dy}{dx}=sec(tan^{-1}x)tan(tan^{-1}x)\times \frac{d}{dx}(tan^{-1})$                $[\frac{d}{dx}secx=secx\cdot tanx]$

$\frac{dy}{dx}=sec(tan^{-1}x)tan(tan^{-1}x)\times (\frac{1}{1+x^2})$                    $[\frac{d}{dx}tan^{-1}x=\frac{1}{1+x^2}]$

$\frac{dy}{dx}=\frac{xy}{1+x^2}$              $[y=sec(tan^{-1}x)]$

Therefore the required solution is $\frac{dy}{dx}=\frac{xy}{1+x^2}$.