Question

derivative of tan inverse (a/x) ​

Answer 1

y = tan^-1(a/x)

dy / dx = {1/(1+ a^2/x^2)}{a/(-x^2)}

dy/dx = - a / x^2 + a^2

hope it helped you

Answer 2

Answer:  $\dfrac{-a}{x^2+a^2}$

Step-by-step explanation:

To find : $\dfrac{d}{dx} tan^{-1}( \dfrac{a}{x} )$

As we know

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

Applying the same we get

$y= tan^{-1}( \dfrac{a}{x} )$

$\dfrac{dy}{dx}=\dfrac{d}{dx} tan^{-1}( \dfrac{a}{x} )= \dfrac{1}{1+(\dfrac{a}{x})^2 } .\dfrac{d}{dx} (\dfrac{a}{x} )\\\\\Rightarrow \dfrac{dy}{dx}= \dfrac{x^2}{x^2+a^2} \times \dfrac{-a}{x^2} \\\\\Rightarrow \dfrac{dy}{dx}= \dfrac{-a}{x^2+a^2}$

Hence, the derivative of  $tan^{-1}( \dfrac{a}{x} )$  is    $\dfrac{-a}{x^2+a^2}$