Finding the derivative of x^3 at x=-2

[dec1ble]

i am having trouble with a fairly easy derivative - and was wondering if someone could show me the steps how to find this?

Find this derivative algebraically

f(x) = x^3 (x cubed) at x= -2

The answer in the back of the book says the derivative is 12 - but I did the work and got 4. Please help!

 

[quasar987]

Thy shalt not forget to multiply by the exponent... i.e.

$$(x^a)' = ax^{a-1}$$

 

[courtrigrad]

Consider the function $$x^4$$ Then $$\frac{dy}{dx} = 4x^{3}$$

Do the same for your function

 

[dextercioby]

Use the definition:Denote the derivative in the point "-2" by D.Then:

$$D=:\lim_{x\rightarrow -2}\frac{f(x)-f(-2)}{x-(-2)}=\lim_{x\rightarrow -2}\frac{x^{3}+8}{x+2}=\lim_{x\rightarrow -2} x^{2}-2x+4 =+12$$

,where i made use of the identity:

$$a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$$

Daniel.

 

[p53ud0 dr34m5]

dextercioby did it through definition of a derivative. if you are lazy like me, you can do it a shorter way!
$$y=x^3~~n=3$$
if you have a polynomial function, and you want to find the derivative of it use the fact that if $y=x^n$, then $\frac{dy}{dx} = nx^{n-1}$.
$$\frac{dy}{dx}=3x^{3-1}=3x^2$$
now, evaluate $3x^2$ at -2.
$$\frac{dy}{dx}=3(-2)^2=3(4)=12$$
that's the lazy way.