# Math Help - Chain Rule Differentiation

1. ## Chain Rule Differentiation

2^(Πsinx)

Seems like I'm having difficulty understanding the a^x = a*ln(a) rule...

2. Originally Posted by Neversh
2^(Πsinx)

Seems like I'm having difficulty understanding the $\color{blue}a^x = a^x*ln(a)$ rule...
Hi Neversh,

$\frac{d}{dx}2^{{\pi}sinx}=\frac{du}{du}\frac{d}{dx }2^{{\pi}sinx}$

$u={\pi}sinx$

$a=2$

$\frac{du}{dx}\frac{d}{du}2^u=({\pi}cosx)2^uln2$

$=({\pi}cosx)2^{{\pi}sinx}ln2$