Why can't you do this when differentiating (sinx)^3 Watch

Jack93o
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for (sinx)^2, its simple, you just apply the chain rule, i.e. let y = (sinx)^2, let u = sinx......and eventually you'll get 2sinxcosx, which is equivalent to sin2x


but why is that I can't use the same method for powers of sinx greater than 2?

take (sinx)^3 for example, apparently I have to convert it into this first: (1/4)(3sinx - sin3x)

but why?

I assumed you could just do: let y = (sinx)^3, let u = sinx, dy/du =......etc
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james22
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You can still use the chain rule the same as you did with (sinx)^2.
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brianeverit
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(Original post by Jack93o)
for (sinx)^2, its simple, you just apply the chain rule, i.e. let y = (sinx)^2, let u = sinx......and eventually you'll get 2sinxcosx, which is equivalent to sin2x


but why is that I can't use the same method for powers of sinx greater than 2?

take (sinx)^3 for example, apparently I have to convert it into this first: (1/4)(3sinx - sin3x)

but why?

I assumed you could just do: let y = (sinx)^3, let u = sinx, dy/du =......etc
General result is \frac{d}{dx}\sin^n x=n\sin^{n-1}x \cos x
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