Question

f(x)=x^3 + x and h be the inverse of f. Calculate h'(10).

Answer 1

I have no idea how to start these.

Answer 2

Basically solve for \(x\) where \[
10 = x^3+x
\]

Answer 3

i would make an educated guess first

Answer 4

Actually, it's a derivative right?

Answer 5

again you are going to use
\[(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}\]

Answer 6

you need first to find \(f^{-1}(10)\) then you need \(f'(that)\) and finally flip it

Answer 7

I get that it equals 301?

Answer 8

but the problem is cooked up so that \(f^{-1}(10)\) should be real easy to find
think of solving \(x^3+x=10\) in your head

Answer 9

why do you set them equal to each other?

Answer 10

slow
not \(f(10)\) but rather \(f^{-1}(10)\) i.e. what would you put in for \(x\) to get \(f(x)=10\)

Answer 11

I thought it was a derivative, I got it.

Answer 12

You can get his formula by using chain rule: \[
f(h(x)) = x \implies [f(h(x)) ]' = 1 \implies f'(h(x))h'(x) = 1 \implies h'(x) = \frac{1}{f'(h(x))}
\]

Answer 13

okay, so I set them equal to each other. and solve for x?

Answer 14

k good
this is what you are using for all of these
\[(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}\]
what you need is
\[(f^{-1})'(10)=\frac{1}{f'(f^{-1}(10))}\] so the first number you have to find is the insidest one
\[f^{-1}(10)\] which is another way of saying solve \(f(x)=10\) or in your case solve \(x^3+x=10\)

Answer 15

solve this in your head
then find \(f'(x)\) and then take \(f'(\text{what you found})\) then flip it

Answer 16

2!

Answer 17

eureka!

Answer 18

now we are up to
\[(f^{-1})'(10)=\frac{1}{f'(f^{-1}(10))}=\frac{1}{f'(2)}\]

Answer 19

your next job is to find \(f'(2)\) which again you can probably do in your head

Answer 20

1?

Answer 21

seems unlikely
\[f(x)=x^3+x\] what is
\[f'(x)\]?

Answer 22

don't you set 2 equal to that?

Answer 23

confused still
lets go back to this \[(f^{-1})'(10)=\frac{1}{f'(f^{-1}(10))}=\frac{1}{f'(2)}\]
the denominator is just \(f'(2)\) has nothing to do with the inverse or with solving
just compute that number

Answer 24

so 12? 1/12

Answer 25

yay
except i think it is 13

Answer 26

yep sorry forgot to add 1

Answer 27

lol
hope this is getting clearer

Answer 28

so the final answer is 13?

Answer 29

\[(f^{-1})'(10)=\frac{1}{f'(f^{-1}(10))}=\frac{1}{f'(2)}=\frac{1}{3\times 2^2+1}=\frac{1}{13}\]

Answer 30

wait, no it is 1/13

Answer 31

one over f prime of f inverse of x

Answer 32

I understand now, it will just take a bit of practice! Thank You so much for helping me! I really appreciate it!

Answer 33

yw
now you will get to
\[\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\cos(\sin^{-1}(x))}\]

Answer 34

one over f prime of f inverse of x
you will want to write this as
\[\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}\]

Answer 35

Wow!

Answer 36

that is if you recall that \(\cos(\sin^{-1}(x))=\sqrt{1-x^2}\)

Answer 37

similarly
\[\frac{d}{dx}\ln(x)=\frac{1}{e^{\ln(x)}}=\frac{1}{x}\]thinking of \(f(x)=e^x, f^{-1}(x)=\ln(x)\)

Answer 38

Just had to do another problem like the on above and I knew how to do it! Thanks again!

Answer 39

yw
good work