Thread: Deriving sinusoidal functions rules.. confusion.

1. Deriving sinusoidal functions rules.. confusion.

I have some confusion relating the chain rule and product rule when differentiating sinusoidal functions compared to when I used the rules comfortably when it was just regular functions. I am aware that the derivative of sinx is cosx, and that the derivative of cosx is -sinx btw.

Example 1:

f(x)=sin(sinx)

Would this be the product rule? If so would it look like:
f'(x)=(cosx)(sinx)+(sinx)(cosx)

however the book answer is f'(x)=[cos(sinx)](cosx)

What am I doing wrong?

Example 2:

f(theta)= (-pi/2 sin)(2theta-pi)

product rule? and would the derivative of -pi/2sin be -pi/2cosx?

2. Originally Posted by kmjt
I have some confusion relating the chain rule and product rule when differentiating sinusoidal functions compared to when I used the rules comfortably when it was just regular functions. I am aware that the derivative of sinx is cosx, and that the derivative of cosx is -sinx btw.

Example 1:

f(x)=sin(sinx)

Would this be the product rule? If so would it look like:
f'(x)=(cosx)(sinx)+(sinx)(cosx)

however the book answer is f'(x)=[cos(sinx)](cosx)

What am I doing wrong?

Example 2:

f(theta)= (-pi/2 sin)(2theta-pi)

product rule? and would the derivative of -pi/2sin be -pi/2cosx?
Dear kmjt,

For example 1 you have to use the chain rule. The product rule can be used when there are to functions multiplied together. But in this case $sin(sinx)$ is a single function.

$\frac{d}{dx}sin(sinx)=\frac{d}{d(sinx)}sin(sinx)\t imes{\frac{d}{dx}sinx}=cos(sinx)\times{cosx}$

P.S; Can you please tell what do you mean by "btw".

3. btw = by the way, sorry aha

4. Originally Posted by kmjt
btw = by the way, sorry aha
Dear kmjt,

Thanks. And do you understand you problem now?

5. Oh nevermind I get it now.. thanks!