Formal Definition of Derivative

**DoQrs** · September 30th 2007, 09:05 AM

I need to use the formal definition of a derivative,
g'(x) = lim h->0 (f(x + h) - f(x)) / h

to solve x^(2/3)...

lim h->0 (f(x+h)^(2/3) - x^(2/3)) / h

I know the answer is obviously (2/3)x^(-1/3) but we're supposed to use the difference quotient as stated above. Any help would be greatly appreciated

**Jhevon** · September 30th 2007, 09:10 AM

Originally Posted by DoQrs

I need to use the formal definition of a derivative,
g'(x) = lim h->0 (f(x + h) - f(x)) / h

to solve x^(2/3)...

lim h->0 (f(x+h)^(2/3) - x^(2/3)) / h

I know the answer is obviously (2/3)x^(-1/3) but we're supposed to use the difference quotient as stated above. Any help would be greatly appreciated

$g(x) = x^{2/3}$

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

$= \lim_{h \to 0} \frac {(x + h)^{2/3} - x^{2/3}}h$

now simplify the top and find the limit. i believe rationalizing the top (multiplying by its conjugate over itself) should work

**red_dog** · September 30th 2007, 09:15 AM

$\displaystyle g'(x)=\lim_{h\to 0}\frac{\sqrt[3]{(x+h)^2}-\sqrt[3]{x^2}}{h}=$
$\displaystyle=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h\left(\sqrt[3]{(x+h)^4}+\sqrt[3]{x^2(x+h)^2}+\sqrt[3]{x^4}\right)}=$
$\displaystyle=\lim_{h\to 0}\frac{h(2x+h)}{h\left(\sqrt[3]{(x+h)^4}+\sqrt[3]{x^2(x+h)^2}+\sqrt[3]{x^4}\right)}=$
$\displaystyle=\frac{2x}{3\sqrt[3]{x^4}}=\frac{2}{3\sqrt[3]{x}}=\frac{2}{3}x^{-\frac{1}{3}}$

**DoQrs** · September 30th 2007, 09:23 AM

Thank you, did you use an online tool to generate the images?

**red_dog** · September 30th 2007, 09:26 AM

We use Latex.
See "Latex Help" category, in the home page on this forum.

**Jhevon** · September 30th 2007, 09:42 AM

Originally Posted by DoQrs

Thank you, did you use an online tool to generate the images?

or see our LaTex tutorial here

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