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Given the function and derivative values in the table below, evaluate $\frac{d}{dx}f^{-1}(3)$

    x: 1   2   3   4   5
 f(x): 4   1   5   2   3
df/dx: 3  -1   4   0  -2

All I know is that the derivative of an inverse is $\frac{1}{f^\prime(f^{-1}(x))}$. Could anyone at least give me hints on how to use the table to my advantage? Thank you!

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The only $x$ here that $f$ sends to $3$ is (blank) and so is equal to $f^{-1}(3)$. Now at this point, the value of the derivative is... –  1015 Feb 8 '13 at 3:44

3 Answers 3

Note,

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

Also we have $f(5)=3$, and thus $f^{-1}(3)=5$,

So $$\lim_{x\to 3}\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(5)}=-\frac{1}{2}$$ Sense $f'(5)=-2$

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many thanks! I understand it so much better now! –  user56852 Feb 8 '13 at 4:08

You're asked to evaluate the derivative of the inverse at 3.

You state a particular formula for the derivative of the inverse.

Now plug in 3.

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Notice that $$f(f^{-1}(x))=x$$ Differentiating using chain rule, we obtain $$f'(f^{-1}(x)).(f^{-1})'(x)=1$$ from which we obtain $$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$$ $f(5)=3$, and so we have $f^{-1}(3)=5$ and $f'(5)=-2$.

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