# How do you differentiate f(x)=cot(sqrt(x-3))  using the chain rule?

Mar 28, 2017

$- {\csc}^{2} \frac{\sqrt{x - 3}}{2 \cdot \sqrt{x - 3}}$
Differentiate the outer function with respect to the inner function (ie $\frac{\mathrm{dy}}{\mathrm{du}}$, where du $= \sqrt{x - 3}$ and multiply with the derivative of the inner function with respect to x. ie $\frac{\mathrm{du}}{\mathrm{dx}}$, where u $= \sqrt{x - 3}$ That is, $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$
The derivative of $\cot \left(x\right)$ with respect to $x$ is $- {\csc}^{2} \left(x\right)$, with respect to $g \left(x\right)$ this becomes $- {\csc}^{2} \left(\sqrt{x - 3}\right)$, the derivative of $\sqrt{x - 3}$ is $0.5 {\left(\sqrt{x - 3}\right)}^{-} \left(\frac{1}{2}\right)$ which is $= \frac{1}{2 \cdot \sqrt{x - 3}}$
Thus, we obtain the answer by multiplying the two, so $- {\csc}^{2} \left(\sqrt{x - 3}\right) \cdot \frac{1}{2 \cdot \sqrt{x - 3}} = - {\csc}^{2} \frac{\sqrt{x - 3}}{2 \cdot \sqrt{x - 3}}$