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<math> f'(x) = \frac{d}{dx} (\sin(x))^n = n(\sin(x))^{n-1}\cos(x) </math>
<math> f'(x) = \frac{d}{dx} (\sin(x))^n = n(\sin(x))^{n-1}\cos(x) </math>
===Proof===
The power rule is simple and elegant to prove with the definition of a derivative:
:<math>f'(x) = \lim_{h\rarr0} \frac{f(x+h)-f(x)}{h}. </math>
Substituting <math> f(x) = x^n </math> gives
:<math>f'(x) = \lim_{h\rarr0} \frac{(x+h)^n-x^n}{h}.</math>
The two polynomials in the numerator can be factored out. It is left as a series, since n can be any integer.
:<math>f'(x) = \lim_{h\rarr0} \frac{((x+h) - x)( (x+h)^{n-1} + (x+h)^{n-2}x + (x+h)^{n-3}x^2 + \dots + (x+h)x^{n-2} + x^{n-1})}{h}. </math>
Then if the first factor is eliminated:
:<math>f'(x) = \lim_{h\rarr0} \frac{(h)( (x+h)^{n-1} + (x+h)^{n-2}x + (x+h)^{n-3}x^2\dots + (x+h)x^{n-2} + x^{n-1})}{h}. </math>
Now, the "h" can be eliminated. This is important, since the denominator cannot go to zero':
:<math>f'(x) = \lim_{h\rarr0} (x+h)^{n-1} + (x+h)^{n-2}x + (x+h)^{n-3}x^2\dots + (x+h)x^{n-2} + x^{n-1}. </math>
With no "h" in the denominator, the limit can be evaluated by letting h=0.
:<math>f'(x) = (x+0)^{n-1} + (x+0)^{n-2}x + (x+0)^{n-3}x^2\dots + (x+0)x^{n-2} + x^{n-1}. </math>:
:<math>f'(x) = x^{n-1} + x^{n-1} + x^{n-1}\dots + x^{n-1} + x^{n-1}. </math>
We see that there are ''n'' terms, so:
:<math>f'(x) = nx^{n-1}. </math>
QED
[[Category:Mathematics]]
[[Category:Mathematics]]