# Thread: Differentiating a rational function

1. ## Differentiating a rational function

Hello

Excuse me if I make any improper use of mathematical experssions because English isn't my mother-tongue and I don't study math in English.

I'm a high school student in the 10th. I have math exam tomorrow and while stuyding I came across this question where I must differentiate a rational function. The function is:

y = 3 / ( (2x+1)^3 * sqrt(2x+1) )

I solved it by first adding a minus, then squaring the denominator, then multiplying by the derivative of the denominator, which is 3.5*(2x+1)^2.5 * 2... This obviously gave me a correct answer...

However, I recall that our teacher once mentioned a shortcut to derive a rational function, but gave us a simpler example. He said, in general:

y = 1/x^n
y' = - n / x^n+1

But if I try to solve it that way, I get:

y' = - 10.5 / (2x+1)^4.5, whilst the correct final derivative has 21 in the numerator. That means that I must multiply it by 2, which is the internal-derivative (dunno how u call it, but I'm referring to the 2x+1) of the denominator!

So is that correct? Can I solve it in the above-mentioned method? If so, has the teacher actually missed that specific detail when he gave us the general law?

Louie.

2. Originally Posted by loui1410
Hello

...
The function is:

y = 3 / ( (2x+1)^3 * sqrt(2x+1) )

...
Hello,

$\displaystyle y = \frac{3}{(2x+1)^2 \cdot \sqrt{2x+1}} = 3 \cdot (2x+1)^{-\frac{7}{2}}$ Use chain rule to derivate:

$\displaystyle y' = 3 \cdot \left( -\frac{7}{2} \right) \cdot (2x+1)^{-\frac{7}{2} - 1} \cdot 2$ which simplifies to:

$\displaystyle y' = -21 \cdot (2x+1)^{-\frac{9}{2}} = -\frac{21}{(2x+1)^4 \cdot \sqrt{2x+1}}$

3. Thanks but finding the derivative was not my question.

4. Hello. This is probably a little late, but... You are asking whether you can apply the 'rule'

$\displaystyle y = \frac{1}{x^n} \Rightarrow y' \equiv \frac{dy}{dx} = \frac{-n}{x^{n+1}}$

to differentiate the function

$\displaystyle y = \frac{3}{(2x + 1)^3 \sqrt{2x + 1}} \equiv 3 (2x + 1)^{-7/2}$ ?

The answer is no! Your function is not of the form $\displaystyle 1 / x^n$; it is of the form $\displaystyle 1 / u^n$ where $\displaystyle u$ is a function of $\displaystyle x$ (here, $\displaystyle u = (2x + 1)$).

In a problem like yours, where $\displaystyle y$ is a function of $\displaystyle u$, and $\displaystyle u$ is a function of $\displaystyle x$, you need to use a theorem called the chain rule. The chain rule states:

$\displaystyle y' \equiv \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$ .

So, you can use your 'rule' to find $\displaystyle dy/du$, but then you must multiply it by $\displaystyle du/dx$ to find the true derivative.

Just to give you the full calculation:

$\displaystyle y = \frac{3}{u^{7/2}} \Rightarrow \frac{dy}{du} = \frac{-3 \cdot (7/2)} {u^{9/2}} ~~;$ (simply using your rule!)

$\displaystyle u = 2x + 1 \Rightarrow \frac{du}{dx} = 2 ~~ ;$

$\displaystyle \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{-3 \cdot (7/2)} {u^{9/2}} \cdot 2 = \frac{-21}{(2x+1)^{9/2}}.$

I hope that helps.