Suppose $ F(x)=f(g(x)) $
- $g(1)=3$
- $g'(1)=4$
- $f'(1)=6$
- $f'(3)=5$
What is $F'(1)$ ?
Suppose $ F(x)=f(g(x)) $
What is $F'(1)$ ?
Hint: By the Chain Rule, $F'(x)=g'(x)f'(g(x))$. Now use the information provided to evaluate the various bits when $x=1$.
$F(x)=f(g(x))$, so $F'(x)=f'(g(x))g'(x)$ and for $x=1$ it is $F'(1)=f'(g(1))g'(1)=f'(3)g'(1)=5\cdot4=20$
$F'(x) = f'(g(x))\cdot g'(x)$ $F'(1) = f'(g(1))\cdot g'(1) = f'(3)\cdot 4 = 5 \cdot 4 = 20$