find the derivative of the function. \n y = (tan^{-1}(5x))^{2} \n y =

Explanation:

Step1: Apply the chain rule

Let ( u = \tan^{-1}(5x) ), then ( y = u^{2} ). By the chain rule ( y'=\frac{dy}{du}\cdot\frac{du}{dx} ). First, ( \frac{dy}{du} = 2u ).

Step2: Find ( \frac{du}{dx} )

For ( u=\tan^{-1}(5x) ), using the formula ( \frac{d}{dx}\tan^{-1}(x)=\frac{1}{1 + x^{2}} ) and the chain rule again. Let ( v = 5x ), then ( u=\tan^{-1}(v) ). ( \frac{du}{dv}=\frac{1}{1 + v^{2}} ) and ( \frac{dv}{dx}=5 ). So ( \frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}=\frac{5}{1+(5x)^{2}}=\frac{5}{1 + 25x^{2}} ).

Step3: Substitute back

Since ( u = \tan^{-1}(5x) ) and ( \frac{dy}{du}=2u ), ( \frac{du}{dx}=\frac{5}{1 + 25x^{2}} ). ( y'=2\tan^{-1}(5x)\cdot\frac{5}{1 + 25x^{2}} ).

Answer:

( y'=\frac{10\tan^{-1}(5x)}{1 + 25x^{2}} )