# Question #e4f55

Dec 27, 2017

Use logarithmic differentiation to obtain $\frac{d}{\mathrm{dx}} \left[{\left(\sin \left(x\right)\right)}^{\cos \left(x\right)}\right]$

$= {\left(\sin \left(x\right)\right)}^{\cos \left(x\right)} \cdot \left(\cos \left(x\right) \cot \left(x\right) - \ln \left(\sin \left(x\right)\right) \sin \left(x\right)\right)$.

#### Explanation:

Let $y = {\left(\sin \left(x\right)\right)}^{\cos \left(x\right)}$. Then $\ln \left(y\right) = \cos \left(x\right) \cdot \ln \left(\sin \left(x\right)\right)$ (since $\ln \left({a}^{b}\right) = b \cdot \ln \left(a\right)$).

Now differentiate both sides of this equation with respect to $x$, keeping in mind that $y$ is a function of $x$ to get

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - \sin \left(x\right) \cdot \ln \left(\sin \left(x\right)\right) + \cos \left(x\right) \cdot \frac{1}{\sin} \left(x\right) \cdot \cos \left(x\right)$.

Multiplying both sides by $y = {\left(\sin \left(x\right)\right)}^{\cos \left(x\right)}$ and rearranging gives the answer

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[{\left(\sin \left(x\right)\right)}^{\cos \left(x\right)}\right]$

$= {\left(\sin \left(x\right)\right)}^{\cos \left(x\right)} \cdot \left(\cos \left(x\right) \cot \left(x\right) - \ln \left(\sin \left(x\right)\right) \sin \left(x\right)\right)$.