Question

What is the derivative of ${{\sin }^{2}}x$?

Answer 1

Hint: To solve the given derivative we will use chain rule of differentiation. Now we know that differentiation of the function $f\left( g\left( x \right) \right)$ is $f'\left( g\left( x \right) \right).g'\left( x \right)$ . Now we know that the differentiation of ${{x}^{n}}$ is $n{{x}^{n-1}}$ and differentiation of sinx is cosx. Hence we can easily find the differentiation of the given function.

Complete step by step solution:
First let us understand the concept of composite functions. Composite functions are functions inside functions. Hence simply we say composition of two functions are called composite functions.
Let us take an example to understand this. Suppose we have two functions $f\left( x \right)=5x+2$ and $g\left( x \right)={{x}^{3}}$ . Now we can find two more functions by composition of the functions.
Hence $f\left( g\left( x \right) \right)=5{{x}^{3}}+2$ and $g\left( f\left( x \right) \right)={{\left( 5x+2 \right)}^{3}}$
Now consider the given function ${{\sin }^{2}}x$
Now we know that the function is a composite function of the form $f\left( g\left( x \right) \right)$ where $f\left( x \right)={{x}^{2}}$ and $g\left( x \right)=\sin x$.
Now we know that the differentiation of composite function of the form $f\left( g\left( x \right) \right)$ is given by chain rule of differentiation which states $\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=f'\left( g\left( x \right) \right)g'\left( x \right)$
Now we know that the differentiation of $\sin x$ is $\cos x$.
Also we know that the differentiation of ${{x}^{n}}$ is given by $n{{x}^{n-1}}$
Hence the differentiation of ${{x}^{2}}$ is $2x$
Hence we have $f'\left( g\left( x \right) \right)=2\sin x$ and $g'\left( x \right)=\cos x$
Hence the differentiation of ${{\sin }^{2}}x$ is
$\Rightarrow \left( 2\sin x \right).\left( \cos x \right)$
Now we know that by the double angle formula $\sin 2x=2\sin x\cos x$
$\Rightarrow \sin 2x$
Hence the derivative of ${{\sin }^{2}}x$ is $\sin 2x$

Note: Now note that we can also solve the problem by first converting the given function in terms of $\cos 2x$ by the formula $\cos 2x=1-2{{\sin }^{2}}x$ . Hence we get ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ . Now we can easily differentiate the given function.