1. Derivatives of Inverse Functions

Okay, so I know how to get derivatives, but I have spent the last hour trying to figure this out and this is really frustrating me . Help please. I don't understand how to get the answers to these two problems. I have the answers but the idea is to understand them which i don't.

f(x) = sin x, - pi/2 is less than or equal to x which is less than or equal to, pi/2 with the real number 1/2

f(x) = x^3 - (4/x) with the real number of 6

my professor did a bad job of explaining how to do these problems please help. I'm studying for an exam right now and this is all that seems to be stumping me

2. The inverse function theorem states that $(f^{-1})'(f(x)) = \frac {1}{f'(x)}$

$f(x) = x^{3} - \frac {4}{x}$

$f'(x) = 3x^2 + \frac {4}{x^2}$

so $(f^{-1})'\Big(x^{3}-\frac{4}{x}\Big) = \frac {1}{3x^{2} + \frac {4}{x^{2}}}$

Since you want $(f^{-1})'(6)$, you need to find the value of x such that $x^{3} - \frac {4}{x} = 6$

x = 2

which means $(f^{-1})'(6) = \frac {1}{3(4) + \frac {4}{4}} = \frac {1}{13}$

3. Originally Posted by sgonzalez90
Okay, so I know how to get derivatives, but I have spent the last hour trying to figure this out and this is really frustrating me . Help please. I don't understand how to get the answers to these two problems. I have the answers but the idea is to understand them which i don't.

f(x) = sin x, - pi/2 is less than or equal to x which is less than or equal to, pi/2 with the real number 1/2

f(x) = x^3 - (4/x) with the real number of 6

my professor did a bad job of explaining how to do these problems please help. I'm studying for an exam right now and this is all that seems to be stumping me
What is the question? Are you asking about finding the derivative of the inverse functions, in the first case where f(x)= 1/2 and the second where f(x)= 6?

If so, you can find the derivative of an inverse function by using the chain rule. The definition of inverse function requires that $f(f^{-1}(x))= x$. Differentiating both sides of that, using the chain rule, gives $f^\prime (f^{-1}(x)) (f^{-1})^\prime (x)= 1$ so that $(f^{-1})^\prime(x) = \frac{1}{f^\prime (f^{-1}(x)}$.

If f(x)= sin(x), then f(x)= 1/2 when sin(x)= 1/2 which occurs when $x= \pi/6$. f'(x)= cos(x) so $f'(\pi/6))= cos(\pi/6)= \sqrt{3}{2}$. $(f^{-1})^\prime (1/2)= 1/(\sqrt{3}/2)= \frac{2}{\sqrt{3}}$.

If $f(x)= x^3+ \frac{1}{x}= x^3+ x^{-1}$, then $f(x)= x^3+ x^{-1}= 6$. But I don't believe that has any simple solution so I don't know at what x to take the derivative.