Derivative of a trig function/chain rule

[andrew.c]

Homework Statement

Differentiate sin^3 2x

Homework Equations

Chain rule.

The Attempt at a Solution

I just need to see if I got the answer right, there aren't any solutions available for this paper, and I need it for the next parts of a question...

I got...
<br /> \begin{align*} <br /> = (sin 2x)^3\\<br /> differentiate\\<br /> = 3 (sin 2x)^2 (2 cos 2x)\\<br /> = 6 cos 2x sin^2 2x\\<br /> \end{align*}<br />

?

 

[dx]

Correct.

 

[andrew.c]

Sweet!
Muchos gracias!

 

[andrew.c]

Another quickie...

Is the derivative of y=xe^{-3x} going to work out to be e^{-3x} (1-3x) ?

 

[Mark44]

Yes.

Also, that would be muchas gracias.

 

[Mark44]

Homework Statement

Differentiate sin^3 2x

Homework Equations

Chain rule.

The Attempt at a Solution

I just need to see if I got the answer right, there aren't any solutions available for this paper, and I need it for the next parts of a question...

I got...
<br /> \begin{align*} <br /> = (sin 2x)^3\\<br /> differentiate\\<br /> = 3 (sin 2x)^2 (2 cos 2x)\\<br /> = 6 cos 2x sin^2 2x\\<br /> \end{align*}<br />

?

Some help with your notation.

d/dx(sin³(2x))
= 3 (sin 2x)^2 (2 cos 2x)
= 6 cos(2x) sin²(2x)

The first line says that we want to differentiate the expression in parentheses. In the second line, we have taken the derivative, using the chain rule. The third line is still the derivative, but in cleaned-up form.