Derivative using first principle definition

[pbonnie]

Homework Statement

Calculate the derivative of f(x) = x^3 - 3x^2

Homework Equations

(f(x+h) - f(x))/h

The Attempt at a Solution

Just wondering if someone can check my solution.
(f(x+h)- f(x))/h= ((x+h)^3- 3(x+h)^2- (x^3-3x^2))/h
= (x^3+ 3x^2 h+3xh^2+ h^3-3(x^2+ 2xh+ h^2 )- x^3+3x^2)/h
= (x^3+3x^2 h+3xh^2+h^3-3x^2-6xh-3h^2-x^2+3x^2)/h
= (x^3+3x^2 h+3xh^2+h^3-6xh-3h^2-x^2)/h
= x3 + 3x2 + 3xh + h2 – 6x – 3h –x2
= x3 + 2x2 + 3xh + h2 – 6x – 3h

lim┬(h→0)⁡〖x^3+ 2x^2- 6x〗

f’(x) = x^3+ 2x^2- 6x

 

[Dick]

Homework Statement

Calculate the derivative of f(x) = x^3 - 3x^2

Homework Equations

(f(x+h) - f(x))/h

The Attempt at a Solution

Just wondering if someone can check my solution.
(f(x+h)- f(x))/h= ((x+h)^3- 3(x+h)^2- (x^3-3x^2))/h
= (x^3+ 3x^2 h+3xh^2+ h^3-3(x^2+ 2xh+ h^2 )- x^3+3x^2)/h
= (x^3+3x^2 h+3xh^2+h^3-3x^2-6xh-3h^2-x^2+3x^2)/h
= (x^3+3x^2 h+3xh^2+h^3-6xh-3h^2-x^2)/h
= x3 + 3x2 + 3xh + h2 – 6x – 3h –x2
= x3 + 2x2 + 3xh + h2 – 6x – 3h

lim┬(h→0)⁡〖x^3+ 2x^2- 6x〗

f’(x) = x^3+ 2x^2- 6x

You are doing the right general thing but the algebra starts going badly after the third line. How come the x^3 and the 3x^2 didn't just cancel? Then I completely start losing track.

 

[pbonnie]

I thought they could only cancel out if they were of the same degree?

 

[pbonnie]

It's 3x^2h so I think I can't cancel it out without a negative 3x^2h?

 

[pbonnie]

Sorry, I copy and pasted it from word and didn't edit it, I see now what you mean

 

[pbonnie]

The third line should read:
(x^3 + 3x^2h + 3xh^2 + h^3 - 3x^2 - 6xh - 3h^2 - x^2 + 3x^2)/h

 

[SammyS]

The third line should read:
(x^3 + 3x^2h + 3xh^2 + h^3 - 3x^2 - 6xh - 3h^2 - x^2 + 3x^2)/h

There's still at least one error/typo there .

Should be :

(x^3 + 3x^2h + 3xh^2 + h^3 - 3x^2 - 6xh - 3h^2 - x^3 + 3x^2)/h

Right ?

 

[pbonnie]

Oh! Typo. So f(x+h) - f(x) should be 3x^2 + 3xh + h^2 - 6x - 3h
And the limit would be
3x^2 - 6x
?

 

[Dick]

Oh! Typo. So f(x+h) - f(x) should be 3x^2 + 3xh + h^2 - 6x - 3h
And the limit would be
3x^2 - 6x
?

That's really (f(x+h)-f(x))/h. But yes, that's the right limit.

 

[pbonnie]

Great, thank you both :)