Differentiate the function f(x)=x^3 in the point a. Use the definition of the derivative for this question. I know that the definition of the derivative is:

$$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

The function $f(x)=x^3$ Now to the the derivative...

I know that $h$ means $\Delta x$. I know that $\Delta x$ means $\Delta x = x_2-x_1 \implies x_2 = x_1 + \Delta x$

I'm just not sure how to obtain the derivative do to the function has an exponent.

Here's my attempt:

$$f'(x) = \lim_{h\to 0}\frac{x^3(x+x^3+x^3)-x^3}{h}$$ Then I got: $$f'(x) = \lim_{h\to 0}\frac{x^3(x + 2 x^3)-x^3}{2x^3}$$ Simplified: $$f'(x) = \lim_{h\to 0}= x$$

I know I did something wrong, because WolframeAlpha says: $\frac{d}{dx}(x^3) = 3 x^2$

I just don't see where exactly I did wrong. (If the problem is obvious and I didn't see it I'm sorry, but I'm currently clueless on how to solve this.)


If $f(x)=x^3,f(x+h)=(x+h)^3,$


So, $$\lim_{h\to 0}\frac{f(x+h)-f(x)}h=\lim_{h\to 0}\frac{h\{(x+h)^2+(x+h)x+x^2\}}h$$

$$=x^2+x^2+x^2$$ as $h\ne0$ as $h\to0$

Alternatively, $f(x+h)-f(x)=(x+h)^3-x^3=x^3+3x^2h+3xh^2+h^3-x^3=h(3x^2+3xh+h^2)$

$$\lim_{h\to 0}\frac{f(x+h)-f(x)}h=\lim_{h\to 0}\frac{h(3x^2+3xh+h^2)}h=3x^2$$ as $h\ne0$ as $h\to0$

  • $\begingroup$ Thank you so much, I see want I did wrong. $\endgroup$ – Kami Mar 18 '13 at 11:06
  • $\begingroup$ @user67254, my pleasure. A practice of $f(x+h)$ based on $f(x)$ can be handy. Like for $f(x)=x^x,x^{\sin x},\cos(\ln x)$ etc. $\endgroup$ – lab bhattacharjee Mar 18 '13 at 11:13

Using the Identity

$$a^3-b^3=(a-b)(a^2+a\cdot b+b^2)$$

We can solve the derivative of


using the definition of derivative

$$f'(x) = \lim_{h\to0}{\frac{f(x+h) - f(x)}{h}}$$ $$=\lim_{h\to0}\frac{(x+h-x)((x+h)^2+(x+h)\cdot(x)+x^2)}{h}$$ $$=\lim_{h\to0}((x+h)^2+(x+h)\cdot(x)+x^2)$$ $$=(x)^2+(x)\cdot(x)+x^2$$ $$=3\cdot x^2$$


Where this $x^3(x+x^3+x^3)$ is coming from? You'd rather substitue $(x+h)$ in place of $x$ when $f$ is to be applied:



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