Question

What is the derivative of \[f\left( x \right) = \sin^{2}\left( x \right) – \cos^{2}\left( x \right)\] ?

Answer 1

Hint: First, we need to differentiate the given expression by applying the derivative power rule and need to be careful in using the power rule.After doing the differentiation, then we need to do some rearrangements of terms and hence we find derivatives of the given expression.
Derivative rules used :
1. Power rule : \[\dfrac{d\left( x^{n} \right)}{{dx}} = nx^{n – 1}\]
2. \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
3. \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]

Complete step by step answer:
Given,
\[f\left( x \right) = \sin^{2}\left( x \right) – \cos^{2}\left( x \right)\]
On differentiating both side,
We get,
\[f^{'}\left( x \right) = \dfrac{d}{{dx}}\left( \sin^{2}\left( x \right) – \cos^{2}\left( x \right) \right)\]
By apply the derivative power rule,
We get,
\[f^{'}\left( x \right) = 2sin^{(2 – 1)}x\dfrac{d}{{dx}}\left( {\sin x} \right) – 2\cos^{(2 – 1)}x\dfrac{d}{{dx}}\left( {\cos x} \right)\]
On simplifying,
We get,
\[f^{'}\left( x \right) = 2\sin x\dfrac{d}{{dx}}\left( {\sin x} \right) – 2\cos x\dfrac{d}{{dx}}\left( {\cos x} \right)\]
On differentiating with the help of derivative rules,
We get,
\[f^{'}(x)\ = 2\sin x\ \cos x - \ 2\cos x\left( - \sin x \right)\]
On simplifying,
We get,
\[f^{'}(x)\ = 2\sin x\ \cos x + 2\sin x\ \cos x\]
On further simplifying ,
We get,
\[f^{'}(x)\ = 4\sin x\ \cos x\]
Thus we get the derivative of \[f\left( x \right) = \sin^{2}\left( x \right) – \cos^{2}\left( x \right)\] is \[4\sin x\ \cos x\] .

Note:
Students should not get confused between the derivative of sin(x) and $\sin^{-1}x$ and the derivative of cos(x) and $\cos^{-1}x$. These functions have similar kinds of derivatives so that may confuse students.