# Thread: Finding Derivative of a Quotient involving Radicals

1. ## Finding Derivative of a Quotient involving Radicals

Alright so the problem simply says "Find the derivative of each of the following functions" and one listed is as follows:

q(x)=((3x-5)^(.5))/((x+4)^(.5))

it is written q(x)= √3x-5 / √x+4

[I'm on a Mac, didn't want to use only symbols in case they show up weird for PC]

2. Originally Posted by Eternal Calculus Student
Alright so the problem simply says "Find the derivative of each of the following functions" and one listed is as follows:

q(x)=((3x-5)^(.5))/((x+4)^(.5))

it is written q(x)= √3x-5 / √x+4

[I'm on a Mac, didn't want to use only symbols in case they show up weird for PC]
$q\!\left(x\right)=\sqrt{\frac{3x-5}{x+4}}$.

I would recommnd using quotient rule in conjunction with chain rule. I'll get you started:

\begin{aligned}q^{\prime}\!\left(x\right)&=\frac{1 }{2}\left(\frac{3x-5}{x+4}\right)^{-1/2}\cdot\frac{\,d}{\,dx}\left[\frac{3x-5}{x+4}\right]\\ &= \frac{1}{2}\sqrt{\frac{x+4}{3x-5}}\cdot\frac{\,d}{\,dx}\left[\frac{3x-5}{x+4}\right]\\ &= \ldots\end{aligned}

Now apply quotient rule and simplify.

Can you take it from here?

3. Oh, definitely, I didn't realize that it would be that simple the other way around. I confused myself doing the quotient rule first @.@ I should be fine now thank you so much

4. Originally Posted by Eternal Calculus Student
Alright so the problem simply says "Find the derivative of each of the following functions" and one listed is as follows:

q(x)=((3x-5)^(.5))/((x+4)^(.5))

it is written q(x)= √3x-5 / √x+4

[I'm on a Mac, didn't want to use only symbols in case they show up weird for PC]
$q(x) = \left(\frac{3x-5}{x+4}\right)^{\frac{1}{2}}$

$q'(x) = \frac{1}{2} \left(\frac{3x-5}{x+4}\right)^{-\frac{1}{2}} \cdot \frac{3(x+4) - (3x-5)}{(x+4)^2}$

you can clean up the algebra ...