# Thread: Simple fraction derivative.

1. ## Simple fraction derivative.

Hi all,

I have been asked to differentiate the function $f(x)=\frac{5}{x}$ where $(x>0)$. I have not been asked previously in exercises (or taught for that matter) how to differentiate fractions like this. I have used Mathcad to find the solution $-\frac{5}{x^2}$ hoping it would reveal a method to me, but with no such luck. I would really appreciate any help on this. I think it may be a rule for powers prior to finding the derivative that I am missing.

Regards,

Shayne.

2. It is just the general power rule for derivatives but you need to use some algebra first.

$f(x)=\frac{5}{x}=5x^{-1} \implies f'(x)=-5x^{-2}=-\frac{5}{x^2}$

You could also use the quotient rule but that would be way too much work.

3. Originally Posted by Hydralisk
Hi all,

I have been asked to differentiate the function $f(x)=\frac{5}{x}$ where $(x>0)$. I have not been asked previously in exercises (or taught for that matter) how to differentiate fractions like this. I have used Mathcad to find the solution $-\frac{5}{x^2}$ hoping it would reveal a method to me, but with no such luck. I would really appreciate any help on this. I think it may be a rule for powers prior to finding the derivative that I am missing.

Regards,

Shayne.
$f(x)=\frac{5}{x} = 5x^{-1}$

By the power rule, the derivative of this is

$(-1)*5x^{-1 -1} = -5x^{-2} = - \frac{5}{x^2}$

4. Originally Posted by TheEmptySet
It is just the general power rule for derivatives but you need to use some algebra first.

$f(x)=\frac{5}{x}=5x^{-1} \implies f'(x)=-5x^{-2}=-\frac{5}{x^2}$
Yeah, what he said!

5. Thanks for that,

I very much appreciate it.

Shayne.

6. I'm not sure I agree that the quotient rule is SO much more work. I often find that it's a bit more calculus up front, but can somtimes be less algebra later. My working:

$\left(\frac{5}{x}\right)'=\frac{(x)(0)-(5)(1)}{(x)^{2}}=-\frac{5}{x^{2}}.$

Where I especially see a savings in effort is when taking derivatives of functions like this:

$\frac{\sin(x)}{\sqrt{1-x^{2}}}.$

If you flip the denominator upstairs and use the product rule, you'll have to turn right around, often, and get a common denominator. Using the quotient rule saves you that step.

My two cents.