Question

Find the ${n^{th}}$ derivative of ${\cos ^2}x$

Answer 1

This is a calculus problem, involving the use of trigonometric functions and their properties. We can solve this using the pattern recognition method wherein we will be differentiating the given function successively first. Afterwards, we can substitute values and solve problems.

Complete answer: The first order derivative is as follows:
${y_1} = \dfrac{d}{{dx}}\left( {{{\cos }^2}x} \right)$
We know $\cos \left( {2x} \right) = 2{\cos ^2}x - 1$
Thus, ${\cos ^2}x = \dfrac{{1 + \cos \left( {2x} \right)}}{2}$
Substituting this value,
${y_1} = \dfrac{d}{{dx}}\left( {\dfrac{{1 + \cos 2x}}{2}} \right) \\\ {y_1} = \dfrac{1}{2}\dfrac{d}{{dx}}\left( 1 \right) + \dfrac{1}{2}\left( {\cos 2x} \right) \\\$
Now, let $2x = u$ and substitute this value in the above equation:
${y_1} = 0 + \dfrac{1}{2}\dfrac{d}{{du}}\left( {\cos u} \right)\dfrac{d}{{dx}}\left( {2x} \right) \\\ {y_1} = - \sin 2x \\\$
Since we now have an initial function of cosine, let us do the conversion.
We know that $- \sin x = \sin \left( { - x} \right)$
And $\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)$
Therefore,
${y_1} = \sin \left( { - 2x} \right) \\\ {y_1} = \cos \left( {\dfrac{\pi }{2} + 2x} \right) \\\$
Let us go for the second derivative now:
${y_1} = \dfrac{d}{{dx}}\cos \left( {\dfrac{\pi }{2} + 2x} \right)$
Let us assume that $\dfrac{\pi }{2} + 2x = v$
Therefore,
${y_2} = \dfrac{d}{{dv}}\cos v\dfrac{d}{{dx}}\left( {\dfrac{\pi }{2} + 2x} \right) \\\ {y_2} = \dfrac{d}{{dv}}\cos v\left[ {\dfrac{d}{{dx}}\left( {\dfrac{\pi }{2}} \right) + \dfrac{d}{{dx}}2x} \right] \\\ {y_2} = - \sin \left( {\dfrac{\pi }{2} + 2x} \right)\left[ {0 + 2} \right] \\\ {y_2} = - 2\sin \left( {\dfrac{\pi }{2} + 2x} \right) \\\ {y_2} = 2\sin \left( { - \left[ {\dfrac{\pi }{2} + 2x} \right]} \right) \\\$
We now convert the functions into cosines again:
${y_2} = 2\cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{2} + 2x} \right) \\\ {y_2} = 2\cos \left( {\dfrac{{2\pi }}{2} + 2x} \right) \\\$
We here notice a pattern:
${y_n} = n\cos \left( {\dfrac{{n\pi }}{2} + 2x} \right)$
This is the ${n^{th}}$ derivative of the given function.
Hence, the ${n^{th}}$ derivative of ${\cos ^2}x$ is $n\cos \left( {\dfrac{{n\pi }}{2} + 2x} \right)$

Note:
This is a very simple problem where you have to take the derivative of the given function again and again and notice the pattern and then generalise it in the context of $n$ . You can compute the third, fourth, fifth and so on derivatives also to verify the answer. However you need not show more than two or three derivatives as it will consume a lot of time.