Find the derivative. \( y=\left(\tan ^{-1} x\right)^{5} \) \( y=\square \)

To find the derivative of \( y = \left(\tan^{-1} x\right)^{5} \), we can use the chain rule. The chain rule tells us to take the derivative of the outer function and multiply it by the derivative of the inner function. First, let \( u = \tan^{-1} x \). Then, \( y = u^5 \). Now, differentiating \( y \) with respect to \( u \) gives us: \[ \frac{dy}{du} = 5u^4 \] Next, we need to find \( \frac{du}{dx} \): \[ \frac{du}{dx} = \frac{1}{1+x^2} \] Now, apply the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 5u^4 \cdot \frac{1}{1+x^2} \] Substituting back \( u = \tan^{-1} x \): \[ \frac{dy}{dx} = 5\left(\tan^{-1} x\right)^{4} \cdot \frac{1}{1+x^2} \] So, the derivative is: \[ y' = \frac{5\left(\tan^{-1} x\right)^{4}}{1+x^2} \]