Question

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

Answer 1

Hint:To solve this problem, first we have to find the first derivative of ${\sin ^2}x$. Then on differentiating it for some more times, we will be able to find a specific trend in the derivatives. The trend may change depending on whether the no. of derivative is odd or even, that is like, if it is the ${2^{nd}}$ derivative, here $2$ is even, and if it is the ${3^{rd}}$ derivative, here $3$ is odd. The ${n^{th}}$ derivative may change depending on whether $n$ is odd or even.

Complete step by step answer:
The function given to us is: $f\left( x \right) = {\sin ^2}x$.
Differentiating the function once w.r.t $x$ (using the chain rule), we get the first derivative as: $f'\left( x \right) = 2\sin x\cos x$
At first glance we may suspect that to gain further derivatives we will require the product rule and their form will become increasingly more complex.However we know that:
$\sin 2A = 2\sin A\cos A$

Thus, using the above property of trigonometry, allowing us to write the first derivative as:
$f'\left( x \right) = \sin 2x$
So, by differentiating the function further times, we get:
${f^{\left( 2 \right)}}\left( x \right) = 2\cos 2x$
$\Rightarrow {f^{\left( 3 \right)}}\left( x \right) = - {2^2}\sin 2x$
$\Rightarrow {f^{\left( 4 \right)}}\left( x \right) = - {2^3}\cos 2x$
$\Rightarrow {f^{\left( 5 \right)}}\left( x \right) = {2^4}\sin 2x$
$\Rightarrow {f^{\left( 6 \right)}}\left( x \right) = {2^5}\cos 2x$
$\Rightarrow {f^{\left( 7 \right)}}\left( x \right) = - {2^6}\sin 2x$
$\Rightarrow {f^{\left( 8 \right)}}\left( x \right) = - {2^7}\cos 2x$
⁞
Here, we can see a clear pattern is now forming in the simultaneous derivatives of the function.Thus by observing the pattern of the derivatives we can conclude that the ${n^{th}}$ derivative is:
\[g(x)=\left\{ \begin{align}
  & (-1)^{\dfrac{n}{2}+1}\times2^{n-1}\cos 2x,n>0, \,\,even \\
 & (-1)^{\dfrac{n+1}{2}}\times2^{n-1}\sin 2x,n>0, \,\,odd \\
\end{align} \right.\]

Note:We can find the ${n^{th}}$ derivative of many functions with the same method. If the ${n^{th}}$ derivative of any function is known, we can find the derivative of the function up to how many times we want just by replacing $n$ with the desired number.Most of the functions either follow a trend in their derivative or reach the value zero after a finite number of derivations.