2^(Πsinx) Seems like I'm having difficulty understanding the a^x = a*ln(a) rule...
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Originally Posted by Neversh 2^(Πsinx) Seems like I'm having difficulty understanding the $\displaystyle \color{blue}a^x = a^x*ln(a)$ rule... Hi Neversh, $\displaystyle \frac{d}{dx}2^{{\pi}sinx}=\frac{du}{du}\frac{d}{dx }2^{{\pi}sinx}$ $\displaystyle u={\pi}sinx$ $\displaystyle a=2$ $\displaystyle \frac{du}{dx}\frac{d}{du}2^u=({\pi}cosx)2^uln2$ $\displaystyle =({\pi}cosx)2^{{\pi}sinx}ln2$
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