Question

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

Answer 1

Hint: 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)