can someone describe to me how to find the derivative of an inverse function?

Question

anonymous · Answer

Question?

anonymous · Answer

Question?

anonymous · Answer

Follow this...

anonymous · Answer

suppose $$y = f(x) = 2x ^{2} - 3x$$ if h(x) is the inverse function of f, then h'(-1) equals...

jamesj · Answer

In general, the derivative of the inverse function is
$$ (f^{-1})'(a) = \frac{1}{f'(f^{-1}(a)} $$

Can you do it now?

anonymous · Answer

i don't understand how to do $$f(x ^{-1})$$

jamesj · Answer

In your notation
$$ f(x) = 2x^2 - 3x $$
and h is the inverse, 
$$ h = f^{-1} $$

Hence
$$ h'(-1) = \frac{1}{f'(h(1))} $$

So you first step is to find h(1).  I.e., for what value(s) of x is f(x) = 1?

anonymous · Answer

oh! okay

mathteacher1729 · Answer

mvkellogg -- do you understand the geometry of this concept?  (how the derivative of a function and its inverse are related)

jamesj · Answer

(there should be a minus sign in my last equation; I'm sure you figured that out already.)

anonymous · Answer

mathteacher1729.. kind of....

mathteacher1729 · Answer

Long story short: 

f(x) has a tangent line of slope m at the point (x,y). 
f inverse has a tangent line of slope (-1/m) at point (y,x). 

I'm going to try to find a good pic of this somewhere.

anonymous · Answer

alright