Find the derivative of $LaTeX: \displaystyle f(x) = \sin{\left(2^{x} \right)}$ .

Decomposing the function gives $LaTeX: \displaystyle f(u) = \sin{\left(u \right)}$ , $LaTeX: \displaystyle u = 2^{v}$ , and $LaTeX: \displaystyle v = x.$ Using the chain rule $LaTeX: \displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx}$ . $LaTeX: \displaystyle f'(x) = (\cos{\left(u \right)})(2^{v} \ln{\left(2 \right)})(1) = 2^{v} \ln{\left(2 \right)} \cos{\left(u \right)}$ . Substituting back in $LaTeX: \displaystyle u$ and $LaTeX: \displaystyle v$ gives $LaTeX: \displaystyle f'(x) = 2^{v} \ln{\left(2 \right)} \cos{\left(2^{v} \right)} = 2^{x} \ln{\left(2 \right)} \cos{\left(2^{x} \right)}$ .