Differentiate the function and find the slope of the tangent line at the given value of the independent variabale.

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

$\displaystyle \underset{h\to 0}{\mathop{\lim }}\frac{f(x + h) - f(x)}{h}$

$\displaystyle \underset{h\to 0}{\mathop{\lim }}\frac{(x + h + 1)^3 - (x + 1)^3}{h}$

$\displaystyle = \frac{(1+x^3+3x^2h+3x^2+3h^2x+6hx+3x+h^3+3h^2+3h) - (x^3+3x^2+3x+1)}{h}$

$\displaystyle =\frac{h(3x^2+3hx+6x+h^2+3h+3)}{h}$

$\displaystyle =3x^2 + 3hx + 6x + h^2 + 3h + 3$

$\displaystyle =3x^2 + 6x + 3$

$\displaystyle y'=3x^2 + 6x + 3 \rightarrow y'=3(-2)^2 + 6(-2) + 3 \rightarrow y'=3$

Correct?