# Differentiating

• Sep 23rd 2009, 10:11 PM
VitaX
Differentiating
Differentiate the function and find the slope of the tangent line at the given value of the independent variabale.

$y = (x + 1)^3$, $x = -2$

$\underset{h\to 0}{\mathop{\lim }}\frac{f(x + h) - f(x)}{h}$
$\underset{h\to 0}{\mathop{\lim }}\frac{(x + h + 1)^3 - (x + 1)^3}{h}$
$= \frac{(1+x^3+3x^2h+3x^2+3h^2x+6hx+3x+h^3+3h^2+3h) - (x^3+3x^2+3x+1)}{h}$
$=\frac{h(3x^2+3hx+6x+h^2+3h+3)}{h}$
$=3x^2 + 3hx + 6x + h^2 + 3h + 3$
$=3x^2 + 6x + 3$
$y'=3x^2 + 6x + 3 \rightarrow y'=3(-2)^2 + 6(-2) + 3 \rightarrow y'=3$

Correct?
• Sep 23rd 2009, 10:22 PM
mr fantastic
Quote:

Originally Posted by VitaX
Differentiate the function and find the slope of the tangent line at the given value of the independent variabale.

$y = (x + 1)^3$, $x = -2$

$\underset{h\to 0}{\mathop{\lim }}\frac{f(x + h) - f(x)}{h}$
$\underset{h\to 0}{\mathop{\lim }}\frac{(x + h + 1)^3 - (x + 1)^3}{h}$
$= \frac{(1+x^3+3x^2h+3x^2+3h^2x+6hx+3x+h^3+3h^2+3h) - (x^3+3x^2+3x+1)}{h}$

Correct route so far?

Yes. But is there any particular reason why you getting the derivative from first principles?
• Sep 23rd 2009, 10:29 PM
VitaX
Quote:

Originally Posted by mr fantastic
Yes. But is there any particular reason why you getting the derivative from first principles?

What do you mean exactly first principles? You mean the route I took as opposed to other means? Well I've only learned this method so far.