Question

Find the derivatives of each of the functions in Exercises 17–50. In some cases it may be convenient to do some preliminary algebra

$f\left(x\right)=x^2\arctan x^2$

Short answer

The derivative is$2x\arctan x^2+x^2\left(\frac{2x}{1+x^4}\right)$.

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)=x^2\arctan x^2$.

Step 2. Preliminary Algebra.

It is known,

$$\begin{gathered} \left(fg{)}^{\text{'}}\right(x)=f^{\text{'}}(x\left)g\right(x)+f(x\left)g^{\text{'}}\right(x\left) \\ \right(f\circ u{)}^{\text{'}}\left(x\right)=f^{\text{'}}\left(u\right(x\left)\right)u^{\text{'}}\left(x\right) \\ \frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2} \end{gathered}$$

Step 3. Derivative of the function.

The derivative of the function will be,

$$\begin{gathered} \frac{d}{dx}\left(x^2\arctan x^2\right)=\left(\frac{d}{dx}\left(x^2\right)\right)\arctan x^2+x^2\left(\frac{d}{dx}\left(\arctan x^2\right)\right) \\ =2x\arctan x^2+x^2\left(\frac{1}{1+{\left(x^2\right)}^2}\frac{d}{dx}\left(x^2\right)\right) \\ 2x\arctan x^2+x^2\left(\frac{2x}{1+x^4}\right) \end{gathered}$$