This can be done either by the Chain Rule ($\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\prime \left(g\left(x\right)\right)\cdot g\prime \left(x\right)$) or by recognizing that $e}^{x\cdot \mathrm{ln}\left(2\right)}={e}^{\mathrm{ln}\left({2}^{x}\right)}={2}^{x$ and recalling that $\frac{d}{dx}\left({b}^{x}\right)=\mathrm{ln}\left(b\right)\cdot {b}^{x}$ when $b>0$.