Differentiate the following functions with respect to $ x $ . (a) $ \sin \left( {{x}^{2}}+5 \right) $ (b) $ \cos \left( \sin x \right) $ \[\]

Hint: We recall the definition of composite function $ gof\left( x \right)=g\left( f\left( x \right) \right) $ . We recall the chain rule of differentiation $ \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} $ where $ y=gof $ and $ u=f\left( x \right) $ . We first find $ u=f\left( x \right) $ as the function inside the bracket and $ y $ as the given function and then differentiate using chain rule.\[\]

Complete step-by-step answer:
If the functions $ f\left( x \right),g\left( x \right) $ are defined within sets $ f:A\to B $ and $ g:B\to C $ then the composite function from A to C is defend as $ g\left( f\left( x \right) \right) $ within sets $ gof:A\to C $ . If we denote $ g\left( f\left( x \right) \right)=y $ and $ f\left( x \right)=u $ then we can differentiate the composite function using chain rule as
\[\dfrac{d}{dx}g\left( f\left( x \right) \right)=\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}\]
(a) We are asked to differentiate the function $ \sin \left( {{x}^{2}}+5 \right) $ . We see that it is a composite function which is made by functions $ \sin x $ and $ {{x}^{2}}+5 $ . Let us assign the function in the bracket as $ f\left( x \right)={{x}^{2}}+5=u $ and $ g\left( x \right)=\sin x $ . So we have $ g\left( f\left( x \right) \right)=g\left( {{x}^{2}}+5 \right)=\sin \left( {{x}^{2}}+5 \right)=y $ . We differentiate using chain rule to have;
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} \\
 & \Rightarrow \dfrac{d}{dx}\sin \left( {{x}^{2}}+5 \right)=\dfrac{d\left( \sin \left( {{x}^{2}}+5 \right) \right)}{d\left( \left( {{x}^{2}}+5 \right) \right)}\times \dfrac{d}{dx}\left( {{x}^{2}}+5 \right) \\
 & \\
\end{align}\]
We use the known differentiation $ \dfrac{d}{du}\left( \sin u \right)=\cos u $ for $ u={{x}^{2}}+5 $ to have;
\[\Rightarrow \dfrac{d}{dx}\sin \left( {{x}^{2}}+5 \right)=\cos \left( {{x}^{2}}+5 \right)\times \dfrac{d}{dx}\left( {{x}^{2}}+5 \right)\]
We use the sum rule of differentiation and have;
\[\Rightarrow \dfrac{d}{dx}\sin \left( {{x}^{2}}+5 \right)=\cos \left( x+5 \right)\times \dfrac{d}{dx}{{x}^{2}}+\dfrac{d}{dx}5\]
We use the differentiation formula $ \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} $ for $ n=2 $ and the information that differentiation of constant like 5 in the above step is zero to have;
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\sin \left( {{x}^{2}}+5 \right)=\cos \left( x+5 \right)\times 2{{x}^{2-1}}+0 \\
 & \Rightarrow \dfrac{d}{dx}\sin \left( {{x}^{2}}+5 \right)=\cos \left( x+5 \right)\times 2x \\
 & \Rightarrow \dfrac{d}{dx}\sin \left( {{x}^{2}}+5 \right)=2x\cos \left( x+5 \right) \\
\end{align}\]

(b) We are asked to differentiate the function $ \cos \left( \sin x \right) $ . We see that it is a composite function which is made by functions $ \cos x $ and $ \sin x $ . Let us assign the function in the bracket as $ f\left( x \right)=\sin x=u $ and $ g\left( x \right)=\cos x $ . So we have $ g\left( f\left( x \right) \right)=g\left( \sin x \right)=\cos \left( \sin x \right)=y $ . We differentiate using chain rule to have;
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} \\
 & \Rightarrow \dfrac{d}{dx}\cos \left( \sin x \right)=\dfrac{d\left( \cos \left( \sin x \right) \right)}{d\left( \sin x \right)}\times \dfrac{d}{dx}\sin x \\
\end{align}\]
We use the known differentiation $ \dfrac{d}{du}\left( \sin u \right)=\cos u $ for $ u=x $ and $ \dfrac{d}{du}\cos u=-\sin u $ for $ u=\sin x $ in the above step to have;
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\cos \left( \sin x \right)=-\sin \left( \sin x \right)\times \cos x \\
 & \Rightarrow \dfrac{d}{dx}\cos \left( \sin x \right)=-\cos x\sin \left( \sin x \right) \\
\end{align}\]

Note: If $ f\left( x \right) $ and $ g\left( x \right) $ are functions are well defined functions then we know from sum rule of differentiation that $ \dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)=\dfrac{d}{dx}\left\{ f\left( x \right)+g\left( x \right) \right\} $ . We note that the domain of given functions has to be compatible for chain rule. Here the domain of all given functions $ \sin x,\cos x,{{x}^{2}}+5 $ is real number set $ \mathsf{\mathbb{R}} $ and hence compatible. The chain rule is also stated as $ {{\left( fog \right)}^{'}}=\left( {{f}^{'}}og \right)\cdot {{g}^{'}} $