Question

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

Answer 1

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;

& \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{aligned}$$ We use the known differentiation $ \dfrac{d}{du}\left( \sin u \right)=\cos u $ for $ u={{x}^{2}}+5 $ to have; $$\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{aligned} & \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{aligned}$$ (b) We are asked to differentiate the function $ \cos \left( \sin x \right) $ . We see that it is a composite function which is made by functions $ \cos x $ and $ \sin x $ . Let us assign the function in the bracket as $ f\left( x \right)=\sin x=u $ and $ g\left( x \right)=\cos x $ . So we have $ g\left( f\left( x \right) \right)=g\left( \sin x \right)=\cos \left( \sin x \right)=y $ . We differentiate using chain rule to have; $$\begin{aligned} & \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{aligned}$$ We use the known differentiation $ \dfrac{d}{du}\left( \sin u \right)=\cos u $ for $ u=x $ and $ \dfrac{d}{du}\cos u=-\sin u $ for $ u=\sin x $ in the above step to have; $$\begin{aligned} & \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{aligned}$$ **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}^{'}} $