Derivative of functions by first principles

[...dans]

find the derivative of the following functions from 1st principles
1) f(x)= -2x^3
f'(x)=lim[h:0,(f(x+h)-F(x))/h]
f(x)=-2x^3;f(x+h)=-2(x+h)^3
f'(x)=lim[h:0,(-2(x+h)^3-2x^3)/h]
=lim[h:0,(-4x^3-6x^2h-6xh^2-2h^2)/h] (not sure if correct)
= ????? what next how to get rid of denomenator h?

 

[tkhunny]

Why are you posting this again?

For f(x) = -2x^3

We have -f(x) = +2x^3

Try to be more careful.

 

[JeffM]

find the derivative of the following functions from 1st principles
1) f(x)= -2x^3
f'(x)=lim[h:0,(f(x+h)-F(x))/h] OK This is the definition of a derivative, but it is just copying.
f(x)=-2x^3;f(x+h)=-2(x+h)^3 And this is your function Good so far, but no real work as yet.
f'(x)=lim[h:0,(-2(x+h)^3-2x^3)/h] And this is the definition of the derivative of your function. No real work as yet.
=lim[h:0,(-4x^3-6x^2h-6xh^2-2h^2)/h] (not sure if correct) HUH? How in the world did you get this.
= ????? what next how to get rid of denomenator h?

Do them this way.

\(\displaystyle f(x) = 2x^3 \implies - f(x) = - 2x^3.\)

\(\displaystyle f(x + h) = 2(x + h)^3 = 2 *\ what?\)

So what does f(x + h) - f(x) =?

Now divide by h. You will find that it cancels out of the denominator and leaves what?

Take the limit.

If you are still having trouble, tell me what you get for answers up to where you are having problems.