Question

Any thoughts @tkhunny

Answer 1

@zepdrix and I tried to solve this one, but it didn't work out
@tkhunny

Answer 2

How do you know it didn't work out?

Answer 3

Because -11 is not one of the answer choices

Answer 4

the answer is 1

Answer 5

damn sure

Answer 6

how?

Answer 7

Just one thought. You missed the easiest problem on the exam.

You should know this:

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

Rewriting in terms of \(g(x) = f^{-1}(x)\;for\;x=0\)

\(\dfrac{d}{dx}g(x) = \dfrac{1}{f'(g(x))}\)

Are you seeing it, yet?

Rephrasing the LHS, and substituting \(x=0\) in the usual slightly incorrect way

\(g'(0) = \dfrac{1}{f'(g(0))}\)

Has it hit you, yet?

Multiplying my the hopefully nonzero denominator on the RHS

\(f'(g(0))g'(0) = 1\) -- Well, THAT looks familiar!

Answer 8

How do you know that \(\dfrac{d}{dx}f^{-1}(x) = \dfrac{1}{f'(f^{-1}(x))}\)?

Answer 9

@tkhunny @phi

Answer 10

A proof of that property should be in your text book

Answer 11

I'm looking for it..

Answer 12

The formula is in the textbook,but there isn't an explaination of why beyond the fact that derivatives of inverse functions are reciprocals of each other

Answer 13

It's somewheres between obvious and totally confusing.

let y=f(x) and x= g(y) where g is the inverse function of f
for a specific (x0, y0) point we know
y0= f(x0) and x0= g(y0)

form the fraction
\[ \frac{g(y)-g(y_0)}{y-y_0} = \frac{x-x_0}{y-y_0} = \frac{1}{\frac{y-y_0} {x-x_0}}\]
now take the limits
\[ g'(y_0) = \lim_{y\rightarrow y_0} \frac{g(y)-g(y_0)}{y-y_0} =\frac{1}{ \lim_{x\rightarrow x_0}\frac{y-y_0} {x-x_0}}= \frac{1}{f'(x_0)}\]

Answer 14

I get confused at the limit part

Answer 15

that is the definition of derivative

Answer 16

no, I mean I get taking the limit/derivative, but I don't get what's after the second = sign

Answer 17

that is supposed to be the limit as x->x0 of the fraction
we can "move" the limit into the denominator because the 1 up top does not change
the bottom is then the definition of a derivative

Answer 18

Okay, I get that now. How do you change it to f'(x0)?

Answer 19

by definition
\[ f'(x_0) = \lim_{x\rightarrow x_0} \frac{y-y_0}{x-x_0} \]

Answer 20

ok, thank you

Answer 21

so the idea is that
f'(x0)*g'(y0) = 1

if y0 is 0, then x0 = g(0)
so we can say
\[ f'(g(0)) \ g'(0)=1 \]