how to get the derivative of 4-x^2?

Question

abb0t · Answer

$$\frac{ d }{ dx }a^n = na^{n-1}$$ where n is some constant.

zehanz · Answer

If you may use the Power Rule: $\left(x^n\right)'=nx^{n-1}$, this is not that hard.
YOu then only have to know that the derivative of a constant (4) is 0. The "-" sign just get carried over.

zehanz · Answer

On the other hand, it could be that you have to find the derivative using the limit-definition. Could you tell me if that is the case?

zehanz · Answer

@abb0t: you'd better change that a to x, or x to a :)

anonymous · Answer

ok I know that 2x^2-1=2x?

zehanz · Answer

You mean: $\left(x^2\right)'=2x^{2-1}=2x^1=2x$, yes, that is right, so what is the final answer?

anonymous · Answer

yes, that's what I meant... for the final answer is what I am stuck at?

zehanz · Answer

It is the minus-sign:  $\left(4-x^2\right)'=0-2x=-2x$

anonymous · Answer

oh ok I see thanks @ZeHanz

zehanz · Answer

YW!

amistre64 · Answer

i would find the derivative by taking the limit as h tends to zero of the difference quotient$$\lim_{h\to~0}\frac{(4-(x+h)^2)-(4-x^2)}{x+h-x}$$