# What is the derivative of 2^sin(pi*x)?

Sep 27, 2015

$\frac{d}{\mathrm{dx}} {2}^{\sin \left(\pi x\right)} = {2}^{\sin \left(\pi x\right)} \cdot \ln 2 \cdot \cos \pi x \cdot \left(\pi\right)$

#### Explanation:

Using the following standard rules of differentiation:

$\frac{d}{\mathrm{dx}} {a}^{u \left(x\right)} = {a}^{u} \cdot \ln a \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \sin u \left(x\right) = \cos u \left(x\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} a {x}^{n} = n a {x}^{n - 1}$

We obtain the following result:

$\frac{d}{\mathrm{dx}} {2}^{\sin \left(\pi x\right)} = {2}^{\sin \left(\pi x\right)} \cdot \ln 2 \cdot \cos \pi x \cdot \left(\pi\right)$

Sep 27, 2015

Recall that:

$\frac{d}{\mathrm{dx}} \left[{a}^{u \left(x\right)}\right] = {a}^{u} \ln a \frac{\mathrm{du}}{\mathrm{dx}}$

Thus, you get:

$\frac{d}{\mathrm{dx}} \left[{2}^{\sin \left(\pi x\right)}\right]$

$= {2}^{\sin \left(\pi x\right)} \cdot \ln 2 \cdot \left[\cos \left(\pi x\right) \cdot \pi\right]$

$= \textcolor{b l u e}{{2}^{\sin \left(\pi x\right)} \ln 2 \cdot \pi \cos \left(\pi x\right)}$

That means two chain rules. Once on $\sin \left(\pi x\right)$ and once on $\pi x$.