1. ## derivative of inverse

if $f(x)=x+\frac{1}{x}$ find $(f^{-1})'(\frac{17}{4})$

I know that i have to use the formula

$[f^{-1}]'(a)=\frac{1}{f'[f^{-1}(a)]}$

but i am unable to find the inverse.

2. $f(x) = x + \frac{1}{x}$

$f'(x) = 1 - \frac{1}{x^{2}}$

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

notice that $f(4) = 4 + \frac{1}{4} = \frac{17}{4}$

so $(f^{-1})'\Big(\frac{17}{4}\Big) = \frac{1}{1- \frac{1}{16}} = \frac{16}{15}$

3. Hello, yoman360!

This has nothing to do with derivatives . . .

If $f(x)\:=\:x+\frac{1}{x}$. find $f^{-1}\left(\frac{17}{4}\right)$
By inspection, we see that: . $\begin{array}{ccc}f(4) &=& \dfrac{17}{4} \\ \\[-3mm] f\left(\dfrac{1}{4}\right) &=& \dfrac{17}{4} \end{array}$

Therefore: . $f^{-1}\left(\frac{17}{4}\right) \;=\;4,\;\frac{1}{4}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If we must find the inverse function first . . .

We have: . $y \;=\;x + \frac{1}{x}$

Switch varoables: . $x \;=\;y + \frac{1}{y}$

Solve for $y\!:\;\;xy \;=\;y^2 + 1 \quad\Rightarrow\quad y^2 - xy + 1 \:=\:0$

Quadratic Formula: . $y \:=\:\frac{x \pm\sqrt{x^2-4}}{2}$

Hence: . $f^{-1}(x) \;=\;\frac{x\pm\sqrt{x^2-4}}{2}$

Therefore: . $f^{-1}\left(\frac{17}{4}\right) \;=\;\frac{\frac{17}{4} \pm\sqrt{\left(\frac{17}{4}\right)^2-4}}{2} \;=\;\frac{\frac{17}{4}\pm\sqrt{\frac{225}{16}}}{2 }$

. . . . . $=\;\frac{\frac{17}{4} \pm\frac{15}{4}}{2} \;=\; \begin{Bmatrix}
\dfrac{\frac{17}{4} + \frac{15}{4}}{2} &=& \dfrac{\frac{32}{4}}{2} &=& \dfrac{8}{2} &=& 4 \\ \\[-3mm]
\dfrac{\frac{17}{4}-\frac{15}{4}}{2} &=& \dfrac{\frac{2}{4}}{2} &=& \dfrac{\frac{1}{2}}{2} &=& \dfrac{1}{4} \end{Bmatrix}$

4. Wait. Are you looking for the derivative of the inverse at 17/4 or just the inverse at 17/4?

5. Originally Posted by Random Variable
Wait. Are you looking for the derivative of the inverse at 17/4 or just the inverse at 17/4?
derivative. I forgot to put the ' in the question

6. Originally Posted by yoman360
derivative. I forgot to put the ' in the question
Then it should be what I posted above.

7. It's actually kind of a bullshit question because the function is not one-to-one, and therefore does not have an inverse. Was there a domain restriction given to make it a 1-1 function?

8. Originally Posted by drumist
It's actually kind of a bullshit question because the function is not one-to-one, and therefore does not have an inverse. Was there a domain restriction given to make it a 1-1 function?
Yeah, that's true.

Depending on the domain restrictions,

$(f^{-1})'(\frac{17}{4}) = \frac{1}{1 + \frac{1}{1/16}} = \frac{1}{17}$

or $(f^{-1})'(\frac{17}{4}) = \frac{16}{15}$

9. Originally Posted by drumist
It's actually kind of a bullshit question because the function is not one-to-one, and therefore does not have an inverse. Was there a domain restriction given to make it a 1-1 function?
Good point.

10. Originally Posted by Random Variable
$f(x) = x + \frac{1}{x}$

$f'(x) = 1 - \frac{1}{x^{2}}$

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

notice that $f(4) = 4 + \frac{1}{4} = \frac{17}{4}$

so $(f^{-1})'\Big(\frac{17}{4}\Big) = \frac{1}{1- \frac{1}{16}} = \frac{16}{15}$
Thanks your answer is correct and has helped me to understand this. Thank you