Help solving a derivative for f(x) = 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.