What is the Derivative of ( Sin ^ 3 ) x ?

Question

We're using the Chain Rule a lot and I'm pretty sure we're suppose to use in this case as well. The question asks for the Derivative of Sin to the power of 3, x. ( ( Sin ^3 ) x )

I ended up with something along the lines of: ( 3 ( cosx ) ^ 2 ) * 3( x ^ 2). It was wrong, and I'm pretty much up the creek without a paddle.

Does anyone know the answer to this, and if so please explain how you came to this?

Thank You.

Answer 1

f(x) = sin^3 x = (sin x)^3
f'(x) = 3(sin x)^2(cos x)

Basically, in the chain rule, do 1 step at a time. First step is to lower the exponent and bring down the original one. The second step is to take the derivative of whatever was being raised to that power - in this case, sin x. As the derivative of sin x is cos x, you just multiply that to the end.

This pattern would continue by taking the derivative of x, and whatever might be in x, etc...hence the chain rule. But as the derivative of x is 1, you're done =)

Answer 2

Hint: The derivative of (sinx)³ = 3(sinx)² * the derivative of sinx.

What you have there (3x²) is the derivative of x³ which is neither here nor there.

Answer 3

y = sin³x and let u = sin x
y = u³ and du/dx = cos x
dy/du = 3u²
dy/dx = (dy/du).(du/dx)
dy/dx = 3u².cos x = 3.sin²x.cosx = 3(1 - cos²x).cosx

Answer 4

It's (3sin^2x) (cosx)

Answer 5

f(x) = sin³x
f'(x) = 3sin²x.cosx