Answer
The equation for the tangent line for $x=\frac{\pi}{4}$ is
$$y=-24x+6\pi+3$$
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 $x_0=\frac{\pi}{4}$ and $f(x)=3\cot^4x$.
$$f'(x)=(3\cot^4x)'=3\cdot4\cot^3x(\cot x)'=12\cot^3x\cdot(-\csc^2x)=-12\cot^3x\csc^2x$$
For $x=\frac{\pi}{4}$ we have:
$$f(0)=3\cot^3\frac{\pi}{4}=3\cdot1^3=3$$
$$f'(0)=-12\cot^3\frac{\pi}{4}\csc^2\frac{\pi}{4}=-12\cdot1^3\cdot(\sqrt2)^2=-24$$
So the equation for the tangent line for $x=\frac{\pi}{4}$ is:
$$y-3=-24(x-\frac{\pi}{4})\Rightarrow y=-24x+6\pi+3$$
