I am currently encountering a math problem that I can't seem to solve on my own and I think it is because I missed the last math lecture.
Usually I am pretty good when it comes to derivatives but this one seems to be my nemesis.
Can somebody maybe help me out?
Thank you guys!
PS Im trying to use the formating system, I hope I'm getting it right.
Let $F(x)=f(x^3)$ and $G(x)=(f(x))^3$. You also know that $a^2=4$, $f(a)=2$, $f'(a)=7$, $f'(a^3)=4$
Find $F'(a)=$
and
$G''(a)=$
From the chain rule $$F'(a) = \left.(f(x^3))'\right|_a = f'(a^3)\cdot \left.(x^3)'\right|_a = f'(a^3)3a^2 = (4)(3)(4)$$
Apply the chain rule to $F$($x$), and get $F$'($x$), then plug in $a$ into $F$'($x$). Do the same for $G$''($x$).
You need to use the chain rule:
$$\Big(f\big(g(x)\big)\Big)' = f'\big(g(x)\big) \cdot g'(x)\quad \text{ and }\quad \Big(g\big(f(x)\big)\Big)' = g'\big(f(x)\big)\cdot f'(x)$$
In your cases, note that you have $g(x)=x^3$. Then use the known values for $a$.
