Calc2 Finding $f'(2)$ from tangent line

Question

So, I have a Calc 2 problem I am stuck on. The tangent line to $h(x)$ at $x = 2$ is $3x - 2$. It says to find $f'(2)$ given $f(x) = -3[h(x)]^2 + 2x + 2$

Any ideas on how to go about this? Thank you!

Answer 1

Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3\cdot2-2 = 4$.

Now, $$f(x) = -3[h(x)]^2+2x+2$$ By chain rule for h(x) $$f'(x) = -3\cdot2\cdot[h(x)]\cdot [h'(x)]+2$$ x=-2 $$f'(-2) = -6\cdot [h(-2)]*[h'(-2)]+2$$ Simplify $$f'(-2) = -6\cdot 4\cdot 3+2 = -70$$

Answer 2

We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 \cdot 4 \cdot 3 + 2 = \boxed{-70} $