MHB Verify that d/dx(ln x)-1/2 using theorem 7

[karush]

Verify that $\dfrac{d}{dx}\ln(x)=\dfrac{1}{x}$ <br>
by applying Theorem 7<br>
Theorem 7 states that: If f is a one-to-one differentiable function with inverse $f^{-1}$ and $f'(f^{-1}(a))$ then the inverse function is differtiable at a and <br>
$$\dfrac{dy}{dx}=\dfrac{1}{\dfrac{dx}{dy}}$$<br>
then<br>
$$f^{-1}(y)=\dfrac{dy}{dx}=1$$<br>
therefore <br>
$$\dfrac{dy}{dx}<br>
=\dfrac{1}{f'(x)}<br>
=\dfrac{1}{\dfrac{dx}{dy}}$$<br>
so<br>
$$\displaystyle f^{-1}{\dfrac{1}{x}}<br>
=\dfrac{1}{f'\left(f^{-1}<br>
\left(\dfrac{1}{x}\right)\right)}<br>
=\dfrac{1}{f'(x)}=\dfrac{1}{x}$$<br>
<br>
ok I kinda got lost on this review question

 

[skeeter]

To be honest, I became a bit lost also.A stab at what I think you’re looking for ...

$y = \ln{x} \implies x = e^y$

$\dfrac{d}{dy}\left[x=e^y \right] \implies \dfrac{dx}{dy} = e^y \implies \dfrac{1}{\frac{dx}{dy}} = \dfrac{1}{e^y} = \dfrac{1}{x} = \dfrac{dy}{dx}$

... if not, just ignore.

 

[karush]

probably... I didn't get the theorem to well

not sure why the <br> shows up it didn't earlierwell our next stuff is Integration by Parts which I have some a little

Mahalo for the help...

 

[HallsofIvy]

Verify that $\dfrac{d}{dx}\ln(x)=\dfrac{1}{x}$ <br>
by applying Theorem 7<br>
Theorem 7 states that: If f is a one-to-one differentiable function with inverse $f^{-1}$ and $f'(f^{-1}(a))$ then the inverse function is differtiable at a and <br>
$$\dfrac{dy}{dx}=\dfrac{1}{\dfrac{dx}{dy}}$$<br>
then<br>
$$f^{-1}(y)=\dfrac{dy}{dx}=1$$<br>

This simply makes no sense! The inverse function of a given function has nothing to do with the derivative! In addition "$f^{-1}(y)$" says that you are going to give a function depending on y but $\dfrac{dy}{dx}$ is a function of x, not y.

I think what you meant to say was that "if $y= f(x)= ln(x)$ then $x= f^{-1}(y)= e^y$". NOW you can say that $\frac{dx}{dy}= e^y$ so that $\frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}= \frac{1}{e^y}= \frac{1}{e^{ln(x)}}= \frac{1}{x}$ because, of course, $e^{ln(x)}= x$.

therefore <br>
$$\dfrac{dy}{dx}<br>
=\dfrac{1}{f'(x)}<br>
=\dfrac{1}{\dfrac{dx}{dy}}$$<br>
so<br>
$$\displaystyle f^{-1}{\dfrac{1}{x}}<br>
=\dfrac{1}{f'\left(f^{-1}<br>
\left(\dfrac{1}{x}\right)\right)}<br>
=\dfrac{1}{f'(x)}=\dfrac{1}{x}$$<br>
<br>
ok I kinda got lost on this review question