How do you differentiate ${{2}^{\sin \pi x}}$?

Hint: Now to differentiate the function we will consider $y={{2}^{\sin \pi x}}$ . Now we will take log on both sides and use the property ${{\log }_{a}}{{b}^{n}}=n{{\log }_{a}}b$ . Now we will solve the differentiation and substitute the value of y. Hence we have the differentiation of the given function.

Complete step-by-step answer:
Now since there is power in expression we will use the logarithm method to find the differentiation.
Now consider $y={{2}^{\sin \pi x}}$ .
Let us take logs on both sides of the expression.
Hence we have, $\ln y=\ln {{2}^{\sin \pi x}}$
Now we have the property of log which says ${{\log }_{a}}{{b}^{n}}=n{{\log }_{a}}b$ . Hence we have,
$\Rightarrow \ln y=\sin \pi x\left( \ln 2 \right)$
Now differentiating the above equation we get,
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\left( \ln 2 \right)\dfrac{d\left( \sin \pi x \right)}{dx}$
Now we know that for composite functions we use chain rule of differentiation. According to chain rule we have differentiation of $f\left( g\left( x \right) \right)$ is given by $f'\left( g\left( x \right) \right).g'\left( x \right)$
Hence using this rule we have $f\left( x \right)=\sin x$ and $g\left( x \right)=\pi x$ .
Now using chain rule we get,
$\Rightarrow \dfrac{d\left( \sin \pi x \right)}{dx}=\left( \pi \right)\cos \pi x$
Now substituting this value we get,
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\left( \ln 2 \right)\pi \cos x$
Now substituting the value of y we have,
$\Rightarrow \dfrac{1}{{{2}^{\sin \pi x}}}\dfrac{dy}{dx}=\pi \ln 2\cos x$
Now taking ${{2}^{\sin \pi x}}$ to RHS we get,
$\Rightarrow \dfrac{dy}{dx}=\pi \ln 2\left( \cos x \right)\left( {{2}^{\sin \pi x}} \right)$
Hence we have the differentiation of the above equation is $\pi \ln 2\left( \cos x \right)\left( {{2}^{\sin \pi x}} \right)$ .

Note: Now note that can avoid using this method to solve the given problem by considering chain rule. Now we know that according to chain rule we have the differentiation of $f\left( g\left( x \right) \right)$ as $f'\left( g\left( x \right) \right)g'\left( x \right)$ . Now taking $f\left( x \right)={{a}^{x}}$ and $g\left( x \right)=\sin x$ and $h\left( x \right)=\pi x$ . Hence we will find the root of $f\left( g\left( h\left( x \right) \right) \right)$ . Hence the required differentiation will be $f'\left( g\left( h\left( x \right) \right) \right).g'\left( h\left( x \right) \right).h'\left( x \right)$ . Also note while using the logarithm method we have variable y in LHS hence we will have to consider $\dfrac{dy}{dx}$ after differentiation, which again nothing but chain rule.