Issue calculating the derivative of a rational function

[greg_rack]

Homework Statement

Find the derivative of:
$$f(x)=\frac{\sqrt{ax}}{\sqrt{ax}-1}$$

Relevant Equations

Theorems of the algebra of derivatives

First, I calculated the derivative of
$$D(\sqrt{ax})=\frac{a}{2\sqrt{ax}}$$
Then, by applying the due theorems, I calculated the deriv of the whole function as follows:
$$
f'(x)=\frac{\frac{a}{2\sqrt{ax}}(\sqrt{ax}-1)-\sqrt{ax}(\frac{a}{2\sqrt{ax}})}{(\sqrt{ax}-1)^2}=
-\frac{a}{2\sqrt{ax}(\sqrt{ax}-1)^2}$$
Which is not the correct result I should get.

 

[member 587159]

Homework Statement:: Find the derivative of:
$$f(x)=\frac{\sqrt{ax}}{\sqrt{ax}-1}$$
Relevant Equations:: Theorems of the algebra of derivatives

First, I calculated the derivative of
$$D(\sqrt{ax})=\frac{a}{2\sqrt{ax}}$$
Then, by applying the due theorems, I calculated the deriv of the whole function as follows:
$$
f'(x)=\frac{\frac{a}{2\sqrt{ax}}(\sqrt{ax}-1)-\sqrt{ax}(\frac{a}{2\sqrt{ax}})}{(\sqrt{ax}-1)^2}=
-\frac{a}{2\sqrt{ax}(\sqrt{ax}-1)^2}$$
Which has nothing to do with the correct result I should get.

Your work and result are entirely correct though:

https://www.wolframalpha.com/input/?i=\sqrt{ax}/(\sqrt{ax}-1)+derivative

Alternatively, you can define $g(x) = \frac{x}{x-1}$ and note that $f(x) = g(\sqrt{ax})$ and invoke the chain rule to calculate the derivative, but your approach is the most natural.

 

[greg_rack]

Your work and result are entirely correct though:

https://www.wolframalpha.com/input/?i=\sqrt{ax}/(\sqrt{ax}-1)+derivative

Alternatively, you can define $g(x) = \frac{x}{x-1}$ and note that $f(x) = g(\sqrt{ax})$ and invoke the chain rule to calculate the derivative, but your approach is the most natural.

Mmm, that's odd but plausible.
Here's what my textbook regards as the correct form:
$$f'(x)=-\frac{\sqrt{ax}}{2x(\sqrt{ax}-1)^2}$$
Is there a way to write my $\frac{a}{\sqrt{ax}}$ term in the form of $\frac{\sqrt{ax}}{x}$?
I've thought about it but didn't find a solution... is simply the textbook wrong?

 

[PeroK]

Which is not the correct result I should get.

It looks right to me too. Another idea is that sometimes these things are easier with a general function:
$$f(x) = \frac{g(x)}{g(x) - 1}$$ $$f'(x) = \frac{g'(x)}{g(x) - 1} - \frac{g(x)g'(x)}{(g(x) - 1)^2} = \frac{-g'(x)}{(g(x) - 1)^2}$$

 

[PeroK]

$$
-\frac{a}{2\sqrt{ax}(\sqrt{ax}-1)^2} =? -\frac{\sqrt{ax}}{2x(\sqrt{ax}-1)^2}$$

 

[PeroK]

Is there a way to write my $\frac{a}{\sqrt{ax}}$ term in the form of $\frac{\sqrt{ax}}{x}$?
I've thought about it but didn't find a solution... is simply the textbook wrong?

Those expressions look very equal to me!

 

[greg_rack]

Guys, nevermind... I simply didn't notice that $\frac{a}{\sqrt{ax}}$(my result) is just the same as $\frac{\sqrt{ax}}{x}$ from the textbook!
Silly me, by the way, thanks for the ideas and help! @PeroK @Math_QED

 

[member 587159]

Guys, nevermind... I simply didn't notice that $\frac{a}{\sqrt{ax}}$(my result) is just the same as $\frac{\sqrt{ax}}{x}$ from the textbook!
Silly me, by the way, thanks for the ideas and help! @PeroK @Math_QED

Yes, some people like to not have squares in the denominator so one simply multiplier both numerator and denominator by the relevant square so that the square disappears in the denominator.