Finding the 99th derivative of $\sin (x)$
$\endgroup$ 2Find ${\frac{d^{99}}{dx^{99}}(\sin (x))}$.
What should I do after this?
${\frac{d}{dx}(\sin (x))=\cos (x)}$
3 Answers
$\begingroup$Hint. One may prove that $$ \frac{d^{n}}{dx^{n}}(\sin x)=\sin\left(x+n\frac \pi2 \right) $$ giving $$ \frac{d^{99}}{dx^{99}}(\sin x)=\sin\left(x+\frac{99\pi}2 \right)=\sin\left(x+48 \pi+\frac{3\pi}2 \right)=-\cos x. $$
$\endgroup$ $\begingroup$$\frac{d^2}{dx^2} \sin x = -\sin x$
$\frac{d^4}{dx^4} \sin x = \sin x$
So you notice that taking the 96'th derivative will be $\sin x$ again. That is because doing the 96'th derivative is the same as doing 4th derivative 24 times and doing the 4th derivative didn't do anything. Now you just have to do 3 more to get 99 from 96.
$\endgroup$ $\begingroup$After differentiating 4 times the result will be sinx again. So after differentiating any multiple of 4 times will give sinx again . 99÷4 yields remainder 3 so the answer will be -cosx ie it is the same as differentiating sinx 3 times
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