# Math Help - trouble finding derivatives

1. ## trouble finding derivatives

I am having trouble solving these problems, and finding the derivatives. I do not need them solved, but just explain the next step (unless otherwise noted):

1) sin(2x)*cos(2x)
' = sin(2x)(-2sin2x)+cos2x(2cos(2x))
- where did the two come from? wouldn't it just be cos(2x)?

2) cot(x)/sin(x) = cos(x)/sin^2(x)
- shouldn't there be a (-) since in the identity, there is one?

3) 4sec^2(x)
- Alright, this one, I do not quite understand what they did. It looks like they took the derivative of the number (they get 8) times the sec, then times the derivative of sec^2.

2. ## Re: trouble finding derivatives

Originally Posted by droidus
I am having trouble solving these problems, and finding the derivatives. I do not need them solved, but just explain the next step (unless otherwise noted):
1) sin(2x)*cos(2x)
' = sin(2x)(-2sin2x)+cos2x(2cos(2x))
- where did the two come from? wouldn't it just be cos(2x)?
You are using the chain rule. Differentiate $\sin()$ then $2x.$

3. ## Re: trouble finding derivatives

Originally Posted by droidus
I am having trouble solving these problems, and finding the derivatives. I do not need them solved, but just explain the next step (unless otherwise noted):

1) sin(2x)*cos(2x)
' = sin(2x)(-2sin2x)+cos2x(2cos(2x))
- where did the two come from? wouldn't it just be cos(2x)?
The same place the "2" in the first term came from- the "2x" in the trig function. The derivative of sin(u), with respect to u, is cos(u) and if u= 2x, du/dx= 2.
$\frac{d(sin(2x)}{dx}= \frac{d(sin(u))}{du}\frac{du}{dx}= cos(u)(2)= 2cos(2x)$

2) cot(x)/sin(x) = cos(x)/sin^2(x)
- shouldn't there be a (-) since in the identity, there is one?
No, why should there be? cot(x)= cos(x)/sin(x) and that sin(x) already in the denominator gives cos(x)/sin^2(x).
(You may be thinking of the "-" in the derivative of cot(x). There is NO derivative here, just an identity. If you are asking about the derivative of cot(x)/sin(x), you haven't differentiated yet. What does the quotient rule give for the derivative of cos(x)/sin^2(x)?

3) 4sec^2(x)
- Alright, this one, I do not quite understand what they did. It looks like they took the derivative of the number (they get 8) times the sec, then times the derivative of sec^2.
No, they did not take the derivative of any "number", they used the chain rule. Letting u= sec(x), this function is $4u^2$, which has derivative, with respect to u, of 4(2u)= 8u= 8 sec(u). Now multiply that by the derivative of u with respect to x, the derivative of sec(x).

4. ## Re: trouble finding derivatives

oh, ok - see for the first one... thanks!