Question

How do you find the ${{50}^{th}}$ derivative of $y=\cos x$?

Answer 1

Hint: We first find some derivatives to form the recurring pattern. We find the closest number of 50 being multiple of 4. Then we complete the rest of the derivatives. We can also follow the iterations to find the derivatives directly as 50 is in the form of $4n+2$.

Complete step by step solution:
We first try to find some number of derivatives of $y=\cos x$ and try to find the pattern of change that happens in the iteration.
We find first order derivative of $y=\cos x$ which gives \[{{y}_{1}}=\dfrac{d}{dx}\left( \cos x \right)=-\sin x\].
Now we again differentiate to get \[{{y}_{2}}=\dfrac{d}{dx}\left( -\sin x \right)=-\cos x\]. This is a second order derivative.
Now we again differentiate to get \[{{y}_{3}}=\dfrac{d}{dx}\left( -\cos x \right)=\sin x\]. This is a third order derivative.
Now we again differentiate to get \[{{y}_{4}}=\dfrac{d}{dx}\left( \sin x \right)=\cos x\]. This is a fourth order derivative.
We got back the main function in the fourth iteration.
So, after every four differentiation we start again with $y=\cos x$.
We need to differentiate 50 times. We find the closest number of 50 that is multiple of 4.
The number is 48 as it has to be less than 50 and can’t cross 50.
Therefore, for ${{49}^{th}}$ derivative we start with the function $y=\cos x$.
So, ${{49}^{th}}$ derivative gives \[{{y}_{49}}=\dfrac{d}{dx}\left( \cos x \right)=-\sin x\].
Now we again differentiate to get \[{{y}_{50}}=\dfrac{d}{dx}\left( -\sin x \right)=-\cos x\]. This is ${{50}^{th}}$order derivative.
Therefore, ${{50}^{th}}$ derivative of $y=\cos x$ is \[{{y}_{50}}=-\cos x\].

Note: We can find the ${{50}^{th}}$ derivative directly where we form the iteration forms in the form of $4n,4n+1,4n+2,4n+3$. Now the respective derivative forms are $\cos x,-\sin x,-\cos x,\sin x$. Now, 50 is in the form $4n+2$ where the ${{50}^{th}}$ derivative of $y=\cos x$ is \[{{y}_{50}}=-\cos x\].