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ankara55t · Sep 17, 2009, 04:28 PM

What are the steps behind differentiating and simplifying the following:

y = sin^3x

In getting y' = 3sin^2xcosx why do we not get cos^2x?

galactus · Sep 18, 2009, 03:50 AM

By using the chain rule.

$\frac{d}{dx}[sin^{3}(x)]=3sin^{2}(x)cos(x)$

The derivative of the inside times the derivative of the outside.

The derivative of the outside is $3sin^{2}(x)$

The derivative of the inside is $cos(x)$

We get the cos(x) by just differentiating sin(x).

Unknown008 · Sep 18, 2009, 10:33 AM

You can also have it with the product rule.

You know that $\frac{d(sin^2(x))}{dx} = 2sin(x)cos(x)$ ?

Then put it like that: $sin^3(x) = sin^2(x)sin(x)$

Use the product rule here;

$\frac{d(sin^2(x)sin(x))}{dx} = sin^2(x)cos(x) + sin(x).2sin(x)cos(x) = sin^2(x)cos(x) + 2sin^2(x)cos(x) = 3sin^2(x)cos(x)$