Question

How do you find the derivative of ${{\sin }^{2}}{{x}^{2}}$ ?

Answer 1

For answering this question we have been asked to find the derivative of the given function ${{\sin }^{2}}{{x}^{2}}$ . We know that we can write the equation simply as ${{\left( \sin {{x}^{2}} \right)}^{2}}$ . We know that the formulae of derivatives are given as $\dfrac{d}{dx}\sin x=-\cos x$ and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ .

Complete step by step answer:
Now considering from the question we need to find the derivative of the given function ${{\sin }^{2}}{{x}^{2}}$.
We know that the equation can be simply written as ${{\left( \sin {{x}^{2}} \right)}^{2}}$ .
Let us assume $\sin {{x}^{2}}=u$ and ${{x}^{2}}=v$ so by using the formulae of derivatives are given as $\dfrac{d}{dx}\sin x=-\cos x$ and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ we will have
$\Rightarrow \dfrac{d}{dx}v=\dfrac{d}{dx}{{x}^{2}}\Rightarrow 2x$ and
$\begin{aligned} & \Rightarrow \dfrac{d}{dx}u=\dfrac{d}{dx}\left( \sin {{x}^{2}} \right) \\\ & \Rightarrow \dfrac{d}{dx}\left( \sin v \right)=\dfrac{d}{dv}\left( \sin v \right)\left( \dfrac{dv}{dx} \right) \\\ & \Rightarrow \left( -\cos v \right)\left( 2x \right) \\\ & \Rightarrow -2x\cos {{x}^{2}} \\\ \end{aligned}$
Now we need to find the derivative of ${{\left( \sin {{x}^{2}} \right)}^{2}}={{u}^{2}}$ using the derivatives
$\Rightarrow \dfrac{d}{dx}u=-2x\cos {{x}^{2}}$ .
By using them we will have
$\begin{aligned} & \dfrac{d}{dx}{{\left( \sin {{x}^{2}} \right)}^{2}}=\dfrac{d}{dx}{{u}^{2}} \\\ & \Rightarrow \dfrac{d}{du}{{u}^{2}}\left( \dfrac{du}{dx} \right)=2u\left( -2x\cos {{x}^{2}} \right) \\\ & \Rightarrow -4xu\cos {{x}^{2}}=-4x\sin {{x}^{2}}\cos {{x}^{2}} \\\ \end{aligned}$
If we observe here we have the simplified version in the form of $\sin 2\theta =2\sin \theta \cos \theta$ so here we can write this simply as
$\begin{aligned} & \dfrac{d}{dx}{{\sin }^{2}}{{x}^{2}}=-4x\sin {{x}^{2}}\cos {{x}^{2}} \\\ & \Rightarrow -2x\sin 2{{x}^{2}} \\\ \end{aligned}$

Hence we can conclude that the derivative of $\dfrac{d}{dx}{{\sin }^{2}}{{x}^{2}}$ is given as $-2x\sin 2{{x}^{2}}$ . This method of solving is known as Chain rule.

Note: We have to be very careful while answering questions of this type. These questions do not require much calculations so there is very less possibility of doing mistakes in questions of this type. While answering questions of this type we should be sure with our calculations and concepts. Similarly we can find the derivative of $\dfrac{d}{dx}{{\cos }^{2}}{{x}^{2}}$ which will give us $2x\sin 2{{x}^{2}}$as the result. Similarly we can find the derivative of any trigonometric ratios.