Question

Help with Calc please! Problem attached, i need to find g'(10).

Answer 1

Were you able to find g(10) ?

Answer 2

yes

Answer 3

Now we need g'(10) ? Hmmmmm thinking D:

Answer 4

So there is this weird formula that connects inverses and derivatives.
I wouldn't be able to explain where the formula comes from, but here it is:\[\Large \Large \left[f^{-1}\right]'(a)\quad=\quad\frac{1}{f'\left[f^{-1}(a)\right]}\]

Answer 5

so i just plug in what i got for g(10) which was 1 into f'(x) which is 3x^2+5?

Answer 6

\[\Large f^{-1}\quad=\quad g\]So using the formula:\[\Large \Large \left[g\right]'(a)\quad=\quad\frac{1}{f'\left[g(a)\right]}\]

Answer 7

\[\Large \Large \left[g\right]'(10)\quad=\quad\frac{1}{f'\left[g(10)\right]}\]

Answer 8

Yah looks like you've got the right idea :)

Answer 9

\[\Large \Large \left[g\right]'(10)\quad=\quad\frac{1}{f'\left[1\right]}\]

Answer 10

ok thanks! i literally had that formula in front of me i just didn't realize i needed to use it!

Answer 11

yay team \c:/