Question

How do you find the derivatives of $y={{\sin }^{2}}x$?

Answer 1

While solving trigonometric functions you must know derivatives of basic trigonometric functions just like derivatives of $\sin x=\cos x$.
In this question, square term so by using this $\dfrac{dy}{dx}=\dfrac{dx}{dy}.\dfrac{dy}{dx}$.
We can find the derivative of $y={{\sin }^{2}}x$

Complete step by step solution:
The given trigonometric function is
$\Rightarrow$ $y={{\sin }^{2}}x$
We have the find derivatives of ${{\sin }^{2}}x$.
Now,
The given function is
$\Rightarrow$ $y={{\sin }^{2}}x$
Consider
$u=\sin x$ ... (1)
But, we have, ${{\sin }^{2}}x$ ... (2)
So, and given is $y={{\sin }^{2}}x$ ... (3)
From (1), (2) and (3)
$\Rightarrow$ $y={{u}^{2}}$
Now,
By chain rule,
We have formula for find derivative
$\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$ ... (4)
$y$ is differentiate with $u$ and ‘$u$’ is differentiate with ‘$x$’ is equal to the ‘$y$’ differentiate with $x$.
Now,
$y$ is differentiate with $'u'$ is equal to the
$\therefore \dfrac{dy}{dx}=2u$ ... (5)
And,
$u$ is differentiate with $x$ is equal to the
$\dfrac{du}{dx}=\cos x$ ... (6)
Hence,
Put the value of (5) and (6) in equation (4)
We get,
$\dfrac{dy}{dx}=2u.cosx$
From equation (1)
$\dfrac{dy}{dx}=2\sin x\cos x\,\,...\,\,\left( 4=\sin x \right)$

Hence the derivative of $y=\sin x$
$\dfrac{dy}{dx}=2\sin x\cos x$

Note: Always make sure that you must know all trigonometric functions and their basic derivatives example derivatives of $\sin \left( x \right)=\cos \left( x \right),\,cos\left( x \right)=-\sin \left( x \right),\,\tan \left( x \right)={{\sec }^{2}}\left( x \right)$ etc. and always must know and do not write wrong formula of
$\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$ ... (chain rule)