# How do you differentiate y=3^(cotx)?

Jul 21, 2016

This is an interesting problem.

There are different strategies of approaching this question, but I will take what I see simplest: use of the chain rule.

Let $y = {3}^{u}$ and $u = \cot x$. We have to differentiate both.

$y = {3}^{u}$

$\ln \left(y\right) = \ln \left({3}^{u}\right)$

$\ln y = u \left(\ln 3\right)$

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{du}}\right) = \ln 3$

$\frac{\mathrm{dy}}{\mathrm{du}} = y \ln 3$

$\frac{\mathrm{dy}}{\mathrm{du}} = {3}^{u} \ln 3$

Now for $u$:

$u = \cos \frac{x}{\sin} x$

By the quotient rule:

$u ' = \frac{- \sin x \times \sin x - \cos x \times \cos x}{\sin x} ^ 2$

$u ' = \frac{- {\sin}^{2} x - {\cos}^{2} x}{{\sin}^{2} x}$

$u ' = \frac{- \left({\sin}^{2} x + {\cos}^{2} x\right)}{{\sin}^{2} x}$

$u ' = \frac{- 1}{\sin} ^ 2 x$

$u ' = - {\csc}^{2} x$

By the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {3}^{u} \ln 3 \times - {\csc}^{2} x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\csc}^{2} x \times {3}^{\cot x} \times \ln 3$

In summary, the derivative of $y = {3}^{\cot \left(x\right)}$ is $\frac{\mathrm{dy}}{\mathrm{dx}} = - {\csc}^{2} x \times {3}^{\cot x} \times \ln 3$.

Hopefully this helps!