0

How do you differentiate $$\frac{x^2}{65x-x^2}$$

I keep getting the wrong answer when I try to do quotient rule. Could someone walk me through it?

saulspatz
  • 53,131
super potato
  • 15
  • 2
  • 1
    Where does your attempt go awry? – Angina Seng Apr 12 '20 at 02:37
  • 1
    Welcome to Math SE. Please update your question text to indicate what you've tried, including your wrong answer, as well as the answer you believe is correct (if you know it), so we can better help you determine what you're doing incorrectly. – John Omielan Apr 12 '20 at 02:38
  • Oh, my bad. I should've said I'm having difficulty collecting like terms at the end. I get (195x^2 - 3x^3) / (65x - x^2)^2. The book is showing a different answer however. – super potato Apr 12 '20 at 02:42
  • @superpotato The numerator in your last comment is incorrect. It's best if you show your work so we can point out your specific mistake. You'll want to use MathJax to format your posts. You'll get a much better response if your posts are easy to read. Don't worry if you don't get all the formatting right at first; someone will fix it for you. – saulspatz Apr 12 '20 at 02:51
  • Oh my bad again. I'll learn about MathJax for the future. Thank you for all of your advice! I appreciate it. – super potato Apr 12 '20 at 02:52

1 Answers1

0

For $x\ne0, x\ne65$, we have:

$$\frac{x^2}{65x-x^2}=\frac{x^2}{x(65-x)}=\frac{x}{65-x}$$

Using the rule of differentiation of quotient, that is:

If $f(x)=\frac{u(x)}{v(x)}$ , then $f'(x)=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ , we get:

$$\frac{(1)(65-x)-(-1)(x)}{(65-x)^2}=\frac{65}{(65-x)^2}$$

as the derivative of the given expression. Hope this helps.

Hussain-Alqatari
  • 5,065
  • 2
  • 13
  • 39
  • 1
    You can also use the fact that $\frac{d}{dx}\left(\frac{x}{65-x}\right)=\frac{d}{dx}\left(\frac{x}{65-x}+1\right)$ to (maybe) simplify it – acat3 Apr 12 '20 at 03:18