OpenStudy (studygurl14):

Any thoughts @tkhunny

2 years ago
OpenStudy (studygurl14):

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

2 years ago
OpenStudy (tkhunny):

How do you know it didn't work out?

2 years ago
OpenStudy (studygurl14):

Because -11 is not one of the answer choices

2 years ago
OpenStudy (saurabh135):

the answer is 1

2 years ago
OpenStudy (saurabh135):

damn sure

2 years ago
OpenStudy (studygurl14):

how?

2 years ago
OpenStudy (tkhunny):

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!

2 years ago
OpenStudy (studygurl14):

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

2 years ago
OpenStudy (studygurl14):

@tkhunny @phi

2 years ago
OpenStudy (phi):

A proof of that property should be in your text book

2 years ago
OpenStudy (studygurl14):

I'm looking for it..

2 years ago
OpenStudy (studygurl14):

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

2 years ago
OpenStudy (phi):

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 \[ g'(y_0) = \lim_{y\rightarrow y_0} \frac{g(y)-g(y_0)}{y-y_0} =\frac{1}{ \lim_{x\rightarrow x_0}\frac{y-y_0} {x-x_0}}= \frac{1}{f'(x_0)}\]

2 years ago
OpenStudy (studygurl14):

I get confused at the limit part

2 years ago
OpenStudy (phi):

that is the definition of derivative

2 years ago
OpenStudy (studygurl14):

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

2 years ago
OpenStudy (phi):

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

2 years ago
OpenStudy (studygurl14):

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

2 years ago
OpenStudy (phi):

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

2 years ago
OpenStudy (studygurl14):

ok, thank you

2 years ago
OpenStudy (phi):

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

2 years ago