1. ## AP practice question

If $f(u) = \sin{u}$ and $u = g(x) = x^2-9$, then $(f \circ g)'(3) =$

I know that $g'(x)= 2x$ but didn't know how to use this,

2. you got this : f composition g = sin(x^2-9) finding it's derivative you get this ..
cos(x^2-9)*2x ,substituting 3 yields cos(9-9)*2*3=1*6=6
gl

3. Originally Posted by bigwave
If $f(u) = \sin{u}$ and $u = g(x) = x^2-9$, then $(f \circ g)'(3) =$

I know that $g'(x)= 2x$ but didn't know how to use this,
$(f \circ g)=\sin{(x^{2}-9)}$
$(f \circ g)' = \cos{(x^{2}-9)} \times 2x$