Differentiating

**VitaX** · September 23rd 2009, 09:11 PM

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?

**mr fantastic** · September 23rd 2009, 09:22 PM

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?

**VitaX** · September 23rd 2009, 09:29 PM

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.

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