Finding Derivatives Using First Principles

[digidako]

Sorry for not showing my work on my last post, I'm on my good for nothing blackberry and apparently cannot type. My question was:
Use first principles definition to find dy/dx:
y=(-4/3x)
lim
x--> 0

---------
(X,-4/3x)
(X+h, -4/3(X+h)

=lim»0 ((-4/3(x+h))-(-4/3x))/x+h-x

I have gotten the answer 4/3x^2 through my proof, but I knew what the answer was (I looked at the back in frustration). I'm afraid I may have broken a few rules when I found a common denominator for the numerator ( (3)(x+h)(3x)) and worked it down to

=lim»0 -12x+12x+12h/[3x+3h](3x)

=lim »0 4h/3x^2h+3hx

Any help is appreciated !

 

[Dick]

Sorry for not showing my work on my last post, I'm on my good for nothing blackberry and apparently cannot type. My question was:
Use first principles definition to find dy/dx:
y=(-4/3x)
lim
x--> 0

---------
(X,-4/3x)
(X+h, -4/3(X+h)

=lim»0 ((-4/3(x+h))-(-4/3x))/x+h-x

I have gotten the answer 4/3x^2 through my proof, but I knew what the answer was (I looked at the back in frustration). I'm afraid I may have broken a few rules when I found a common denominator for the numerator ( (3)(x+h)(3x)) and worked it down to

=lim»0 -12x+12x+12h/[3x+3h](3x)

=lim »0 4h/3x^2h+3hx

Any help is appreciated !

That's still pretty hard to read. You are leaving out too many parentheses. If you mean y=(-4)/(3x) then if you got (4h)/(3x^2+3xh) for the numerator then it's ok. Now the denominator is h, so divide by that and let h->0. 4/(3x^2) is correct.