Thread: Derivative of Sin function Problem

If f(x)= sin²(3-x), then dy/dx(0)= ?

A. -2cos3
B. -2sin3cos3
C. 6cos3
D. 2sin3cos3
E. 6sin3cos3

This is a questions from my AP Calculus BC class, and I can't seem to get the correct answer. Any Thoughts?

Thanks!

Originally Posted by r2d2

If f(x)= sin²(3-x), then dy/dx(0)= ?
Any Thoughts?

Thanks!

Sure, start by differentiating $\displaystyle f(x)$.

It may be the case that you are confused because the problem contains two forms of notation.

Understand that $\displaystyle f'(x)=\frac{dy}{dx}$. The fact that this question would mix and match notation is a strong indication that your teachers are crap. But, don't let it get ya down.

So, you are asked to find $\displaystyle f'(0)$...

so f'(x)= sin²(3-x)= sin(3-x) * (sin(3-x) = cos(3-x) * cos(3-x)

Then plug in f'(0) to get cos(3) * cos(3)?

Would that be the way to go about this problem? I think I am missing an important trigonometric derivative point.... Any Thoughts?

Rewrite as (sin(3-x))²

Then use chain rule:

2(sin(3-x)) (cos(3-x)) (-1)