# Thread: Piece Wise - Limits

1. ## Piece Wise - Limits

Find the Derivative of
Squareroot (x^3)

By the limit defination of the derivative. Not the easy shortcut way. :/

Can anyone point me in the right direction as to how I can start?

I know the derivative is
3/2 * x^1/2 (shortcut way) but can anyone show me the long way?

2. $\lim_{\Delta x\rightarrow 0}f(\frac{(\Delta x+x)-f(x)}{\Delta x})$

$\frac{\sqrt{({\Delta x+x})^{3}}-\sqrt{{x}^{3}}}{\Delta x}$

Then keep solving

3. Hello, EliteNewbz!

Find the derivative by the limit definition of the derivative:

. . . $f(x) \:=\:\sqrt{x^3}$
We know the definition: . $f'(x) \;=\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h}$

But I've never met any teacher who says this is a four-step procedure,
. . and deliberately makes four steps. . (Am I the only one?)

[1] Find $f(x+h)$ . . . Replace $x$ with $x+h$ . . . and simplify.

[2] Subtract $f(x)$ . . . Subtract the original function . . . and simplify.

[3] Divide by $h$ . . . Factor and reduce.

[4] Take the limit . . . Let $h \to 0.$

We have: . $f(x) \:=\:x^{\frac{3}{2}}$

[1] $f(x+h) \:=\:(x+h)^{\frac{3}{2}}$

[2] $f(x+h) - f(x) \;=\;(x+h)^{\frac{3}{2}} - x^{\frac{3}{2}}$

. . Multiply by $\frac{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}$

. . $f(x+h)-f(x)\;=\;\frac{(x+h)^{\frac{3}{2}} - x^{\frac{3}{2}}}{1}\cdot\frac{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} {(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{(x+h)^3 - x^3}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}$

. . $f(x+h) - f(x) \;=\;\frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{3x^2h + 3xh^2 + h^3}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}$

[3] $\frac{f(x+h) - f(x)}{h} \;=\;\frac{3x^2h + 3xh^2 + h^3}{h\left[(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}\right]} \;=\;\frac{h\left(3x^2 + 3xh + h^2\right)}{h\left[(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}\right]}$ . $=\;\frac{3x^2 + 3xh + h^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}$

[4] $\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;\lim_{h\to0}\frac{3x^2 + 3xh + h^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{3x^2 + 0 + 0}{x^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{3x^2}{2x^{\frac{3}{2}}} \;=\;\frac{3}{2}x^{\frac{1}{2}}$

Therefore: . $\frac{d}{dx}\left(\sqrt{x^3}\right) \;=\;\frac{3}{2}x^{\frac{1}{2}}$