Question

Help with calculus please? :)

Answer 1

If f(x) = (1/2)x^3 -x +5
and g(x) = f-1(x), find g'(7)
note f(2) = 7

Answer 2

\[
f(x) = \frac 12 x^3-x+5
\]

Answer 3

\[
g(f(x)) = x
\]

Answer 4

Do you know implicit differentiation?

Answer 5

Yes I do

Answer 6

Hold on on second.

Answer 7

Ok. To be honest though I'm not sure where you're going with g'(x)f'(x)

Answer 8

I wrote it incorrectly anyway.

Answer 9

Okay anyway, does this make sense: \[
g(f(x)) = x
\]

Answer 10

Mhm

Answer 11

First, to make it simpler we let \(y=f(x)\). So it becomes: \[
g(y)=x
\]Does this make sense so far?

Answer 12

Yup

Answer 13

Now can you differentiate both sides with respect to \(x\)?

Answer 14

So g'(y)y' = 1
?

Answer 15

Yes.

Answer 16

Then am i supposed to plug in y and y'?

Answer 17

Yes, so first of all, the problem asks for \(g'(7)\).

Answer 18

This means we have to let \(y=7\).

Answer 19

\[
y=7=f(x) \implies x=2
\]This is due to \(f(2) = 7\).

Answer 20

And \(y'\) is asking for \(f'(2)\).

Answer 21

So we plug in 7 for y and 5 for y'?
since y'(2) is 5?

Answer 22

Yes, assuming \(f'(2) = 5\), then plug those in.

Answer 23

I got g'(7) * (5) = 1

so g'(7) = 1/5

Answer 24

Looks good.

Answer 25

Thanks for your help! much apreciated

Answer 26

To be even more general...

Answer 27

If we thought of \(g\) as the original function, and \(y\) as the inverse function, we could say: \[
y' = \frac{1}{g'(y)} \implies f'(x) = \frac{1}{g'(f(x))}
\]

Answer 28

By symmetry: \[
g'(x) = \frac{1}{f'(g(x))}
\]

Answer 29

Then when we say: \(g'(7)\), we know \(g(7)=f^{-1}(7)=2\).\[
g'(7) = \frac{1}{f'(2)}
\]

Answer 30

yeah that makes sense too

Answer 31

I am not a fan of \(f'(g(x))\) because it is ambiguous whether we mean a derivative with respect to \(x\) with \(g(x)\) plugged in, or a derivative with respect to \(g(x)\).