find the derivative of the function. y = (tan^(-1)(6x))^2 y =

Explanation:

Step1: Apply chain - rule

Let $u = \tan^{-1}(6x)$, so $y = u^{2}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$. Since $y = u^{2}$, then $\frac{dy}{du}=2u$.

Step2: Find $\frac{du}{dx}$

We know that the derivative of $\tan^{-1}(t)$ with respect to $t$ is $\frac{1}{1 + t^{2}}$. Here $t = 6x$, so by the chain - rule again, $\frac{du}{dx}=\frac{d}{dx}(\tan^{-1}(6x))=\frac{6}{1+(6x)^{2}}$.

Step3: Substitute $u$ and $\frac{du}{dx}$ into $\frac{dy}{dx}$

Since $u=\tan^{-1}(6x)$ and $\frac{dy}{du}=2u$, $\frac{du}{dx}=\frac{6}{1 + 36x^{2}}$, then $\frac{dy}{dx}=2\tan^{-1}(6x)\cdot\frac{6}{1 + 36x^{2}}$.

Answer:

$\frac{12\tan^{-1}(6x)}{1 + 36x^{2}}$