Question

Check my answer please?

Answer 1

if h is the inverse function of f and if f(x) = 1/x, then h'(3) =

f'(x) = -x^-2

h'(x) = -x^2

h'(3) = -9

My book had a different answer in the back that confused me

Answer 2

inverse of f(x) = 1/x
is ?

Answer 3

first find h(x)
which is inverse of f(x)

Answer 4

The invese of 1/x is 1/x right?

Answer 5

yes

Answer 6

its derivative is ?

Answer 7

-x^(-2)
right ?

Answer 8

ok so it's just -x^-2 so -1/9

Answer 9

is 1 your answer?

Answer 10

h'(x) = -x^(-2)

h'(3) = -1/9
yes

Answer 11

your book had -1/9 as the answer ?

Answer 12

Yes it did

Answer 13

I think that:
h(x)=1/f(x), so:
\[h'(x)=-\frac{ f'(x) }{ f ^{2} }\]

Answer 14

I thought since the derivative of a function is dy/dx and the derivative of its inverse is dx/dy, that they would be reciprocals of each other. But apparently not...

Answer 15

That makes sense too, @Michele_Laino

Answer 16

thanks! @Jhannybean

Answer 17

'derivative of its inverse is dx/dy,' <<NO
see what Michele posted

Answer 18

I can't see how h(x) = 1/f(x)...

Answer 19

\(\Large (f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))}\)

Answer 20

h(x) is NOT 1/f(x)

Answer 21

h(x) is inverse of f(x) = f^(-1)(x)

its not reciprocal

Answer 22

sorry f I write:
\[h=\frac{ 1 }{ f }=\frac{ 1 }{ 1/x }=x\]
which is the definition of f^-1

Answer 23

First describe what the inverse is..maybe then clarify the inverse as h(x)?

Answer 24

I get it now. Inverse first and then derivative

Answer 25

^^

Answer 26

\[h(f(x))=\frac{ 1 }{ f(x) }=\frac{ 1 }{ 1/x }=x\]

Answer 27

are there still any doubts ?

Answer 28

\[y=\frac{1}{x}\]\[x=\frac{1}{y}\implies f^{-1}(x) =\frac{1}{x}=h(x) ~~...\sf \text{because it's annoying to write inverse derivatives} \]\[h(x) = \frac{1}{x} \implies h(x)=x^{-1}\]\[h'(x) = -\frac{1}{x^2}\]\[h'(3) = -\frac{1}{(3)^2} = -\frac {1}{9}\]

Answer 29

No doubts remaining. Gracias

Answer 30

Just writing out how I thought of it, hahaha.

Answer 31

\[h(f(x))=\frac{ 1 }{ f(x) }=\frac{ 1 }{ 1/x }=x\]

\[\implies h'(f(x)) \cdot f'(x) = 1\]

\[\implies h'(f(x)) = \dfrac{1}{f'(x)}\]

Answer 32

this formula is very useful when finding the inverse is hard

Answer 33

ya it looks that way. I'll make sure to memorize it, thank u

Answer 34

i like the explanation here
http://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/inverseDeriv.html