Derivative of a quotient involving cot(x)

Apply the quotient rule: $h'(x) = \frac{(1+x)\cdot\frac{d}{dx}[\cot x] - \cot(x)\cdot\frac{d}{dx}[1+x]}{(1+x)^2}.$ Since $\frac{d}{dx}[\cot x] = -\csc^2 x$ and $\frac{d}{dx}[1+x] = 1$, this becomes $h'(x) = \frac{-(1+x)\csc^2(x) - \cot(x)}{(1+x)^2}.$

Now substitute $x = \pi/4$. Recall $\cot(\pi/4) = 1$ and $\csc(\pi/4) = \sqrt{2}$, so $\csc^2(\pi/4) = 2$. Plugging in: $h'(\pi/4) = \frac{-(1+\pi/4)\cdot 2 - 1}{(1+\pi/4)^2}.$ Distributing the $-2$ gives $-2 - \pi/2 - 1 = -3 - \pi/2$, so $h'(\pi/4) = \frac{-3-\pi/2}{(1+\pi/4)^2} \approx -1.47.$