Question

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 find the derivative...
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 the answer is wrong, I just cannot figure out what I did wrong.

Answer 1

nonononoonnoo *joker voice*

Answer 2

take the definition of derivative and put it IN YOUR FUNCTION

Answer 3

\[f \prime (x) = \lim_{h \rightarrow 0} \frac{ (x+h)^3 - (x)^3}{ h }\]

Answer 4

By the power rule, the answer is \(3x^2\). However, the definition:
\[lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\
lim_{h\to 0}\frac{(x+h)^3-x^3}{h} \\
lim_{h\to 0}\frac{h^3+3 h^2 x+3 h x^2+x^3-x^3}{h} \\
lim_{h\to 0}\frac{h^3+3 h^2 x+3 h x^2}{h} \\
lim_{h\to 0}h^2+3 h x+3 x^2 \\
3x^2\]

Answer 5

By the way, if you'd like to expand a binomial easily, use the binomial theorem.

Answer 6

Oh. I see what I did wrong now *facedesk*.
\[f'(x) = \lim_{h\to0}{\frac{f(x+h) - f(x)}{h}}\]
\[=\lim_{h\to0}\frac{(x+h-x)((x+h)^2+(x+h)\cdot(x)+x^2)}{h}\]
\[=\lim_{h\to0}((x+h)^2+(x+h)\cdot(x)+x^2)\]
\[=(x)^2+(x)\cdot(x)+x^2\]
\[3x^2\]