Answer
The equation for the tangent line at $x=\sqrt\pi$ is
$$y=8\sqrt\pi x-8\pi$$
Work Step by Step
The equation for the tangent line at $x_0$ is:
$$y-f(x_0)=f'(x_0)(x-x_0)$$
Here is $x_0=\sqrt\pi$ and $f(x)=\tan(4x^2)$.
$$f'(x)=(\tan(4x^2))'=\sec^2(4x^2)(4x^2)'=\sec^2(4x^2)\cdot8x=8x\sec^2(4x^2)$$
For $x=\sqrt\pi$ we have:
$$f(\sqrt\pi)=\tan(4(\sqrt\pi^2)=\tan4\pi=0$$
$$f'(\sqrt\pi)=8\sqrt\pi\sec^2(4(\sqrt\pi)^2)=8\sqrt\pi\sec^2(4\pi)=8\sqrt\pi$$
So, the equation for the tangent line at $x=\sqrt\pi$ is:
$$y-0=8\sqrt\pi(x-\sqrt\pi)\Rightarrow y=8\sqrt\pi x-8\pi$$
