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Find $$\frac{\mathrm{d}^{950}}{\mathrm{d}x^{950}}(\sin x)$$

How are really high-order derivatives found? My try:

$$\frac{\mathrm{d}^{950}}{\mathrm{d}x^{950}}(\sin x)= 950(\cos x)^{949}$$

This doesn't seem right. Help please.

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3 Answers

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Hint

Prove that $$\frac{d^k}{dx^k}\sin x=\sin\left(x+k\frac{\pi}{2}\right)$$

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You're asked to find the 950th derivative of $f(x) = \sin x$.

Observe that $$f'(x) = \cos x, f''(x) = - \sin x, f'''(x) = - \cos x, f^{(4)}(x) = \sin x = f(x).$$

So $f(x) = f^{(4)}(x) = f^{(8)}(x) = ... = f^{(948)}(x) = ... f^{(4n)}(x)$, with $n$ a positive integer.

Extending this, $f^{(k)}(x) = f^{(4n+k)}(x),$ with $n$ a positive integer and $0 \leq k < 4.$

Then,

$$f^{(950)}(x) = f^{(2)}(x) = f''(x) = - \sin x.$$

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The notation $\frac {d^{950}}{dx^{950}} \sin x$ usually means , take the $950$th derivative of $\sin x$ , i.e., differentiate the function $\sin x \;950$ times. But notice that the derivatives are periodic, i.e., they repeat after a certain number of times. $d/dx(\sin x)=\cos x; d/dx(\cos x)=-\sin x \dots$

Then $950$ goes on cycles of $4$ a certain number of times, with some left. How many?

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