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Question

Question: How do you find the 108th derivative of \[y=\cos \left( x \right)\]?...

How do you find the 108th derivative of y=cos(x)y=\cos \left( x \right)?

Explanation

Solution

Now we will differentiate the function again and again till we see a pattern in the differential obtained. Now with the help of this pattern we can see that every 4nth4{{n}^{th}} derivative of the function the cycle repeats as the fourth derivative is cosx\cos x . Hence we can easily find the 108th{{108}^{th}} derivative of the function.

Complete step by step solution:
Now consider the function y=cosxy=\cos x . Now we know that the derivative of cosx\cos x is sinx-\sin x .
Hence we can say that the first derivative of the function y=cos(x)y=\cos \left( x \right) is sinx-\sin x .
Now the derivative of sinx\sin x is nothing but cosx\cos x .
Hence we can say that the second derivative of the function y=cos(x)y=\cos \left( x \right) is cosx-\cos x .
Now again differentiating the function we get the third derivative of the function y=cos(x)y=\cos \left( x \right) as sinx\sin x .
And finally again differentiating the function we get the fourth derivative of the function y=cos(x)y=\cos \left( x \right) as cosx\cos x . Hence we can see that the derivatives of cosx\cos x repeats a cycle after every fourth derivative.
Hence after differentiating the function 4n times we will still get cosx\cos x
Now we want to differentiate the function 108 times.
Now we know that 108 is nothing but 27×427\times 4 .
Hence we want to differentiate the function 4n times where n = 27.

Now we know that on differentiating the function 4 times we get cosx. Hence the 108th{{108}^{th}} derivative of cosx\cos x is nothing but cosx\cos x.

Note: Now note that if the nth{{n}^{th}} derivative is not divisible by 4 then we write n as 4k + r. where r is the remainder obtained after dividing by 4. Hence the nth{{n}^{th}} derivative is same as rth{{r}^{th}} derivative and we can calculate rth{{r}^{th}} derivative as r is less than 4.