Calc 1 question

find the derivitive f(x) = -4/x^2 use the limit process

Using fx=(x+Dx)-fx/Dx

I get -4/(x+Dx)^2/x^2 + 4/x^2 and get lost in the math - am I on the right track?

Hello,

Quote:

---

Originally Posted by weezie23

find the derivitive f(x) = -4/x^2 use the limit process

Using fx=(x+Dx)-fx/Dx

I get -4/(x+Dx)^2/x^2 + 4/x^2 and get lost in the math - am I on the right track?

---

:eek:


$\displaystyle f'(x)=\frac{f(x+Dx)-f(x)}{Dx}$

$\displaystyle f({\color{red}x+Dx})=\frac{-4}{({\color{red}x+Dx})^2}$


---> $\displaystyle f'(x)=\frac{\frac{-4}{(x+Dx)^2}+\frac{4}{x^2}}{Dx}$

Did you catch your mistake ? Now, try to simplify :) (show your work, it'd be better)

It's time to get UNlost. You just have to sort through it.

1) Fix your notation:

These are NOT the same:

x+3/5 and (x+3)/5

Remember your Order of Operations.

2) Don't write things that make no sense.

You have: fx=(x+Dx)-fx/Dx

Notice how "fx" is on both sides. I'm sure you didn't mean that. Plus, there's and 'f' missing on the right -- f(x+Dx)?

As far as I get is 8x^2+8xDx+4Dx^2/(x+Dx)-x^2(Angry)

Quote:

---

Originally Posted by Moo

Hello,



:eek:


$\displaystyle f'(x)=\frac{f(x+Dx)-f(x)}{Dx}$

$\displaystyle f({\color{red}x+Dx})=\frac{-4}{({\color{red}x+Dx})^2}$


---> $\displaystyle f'(x)=\frac{\frac{-4}{(x+Dx)^2}+\frac{4}{x^2}}{Dx}$

Did you catch your mistake ? Now, try to simplify :) (show your work, it'd be better)

---

You didn't listen to me about the notation, did you?

If you are VERY CAREFUL, you will see it. If you are sloppy, it will work only by luck.