Question about a derivative problem

**leezgo** · September 13th 2012, 04:48 PM

The problem is simply

f(x) = x^(3/2)

using (f(x+h) - f(x)) / h and multiply with the conjugate I get

(x+h)^3 - x^3
/
h(x+h)^(3/2) + z^(3/2)

How do I get forward from this? I have been trying for hours now, but the algebra gets really ugly. Is there an easier way to reach the answer?

Sorry I don't know how to write formulas in LaTex

**pickslides** · September 13th 2012, 05:28 PM

Consider $\displaystyle \frac{ (x+h)^{\frac{3}{2}}- x^{\frac{3}{2} }}{h}= \frac{\sqrt{x+h}(x+h)- x\sqrt{x}}{h} = \frac{x\sqrt{x+h}+h\sqrt{x+h}- x\sqrt{x}}{h}$

**Soroban** · September 13th 2012, 08:23 PM

Hello, leezgo!

Differentiate: $f(x) \,=\, x^{\frac{3}{2}}$

Using $\frac{f(x+h) - f(x)}{h}$ and multiply with the conjugate,

. . I get : . $\frac{(x+h)^3 - x^3}{h\big[(x+h)^{\frac{3}{2}}+ x^{\frac{3}{2}}\big]}$ . Good!

How do I get forward from this? . Keep going . . .

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

. . . . . . . . . . . . . $=\;\frac{h(3x^2 + 3xh + h^2)}{h\big[(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}\big]} \;=\;\frac{3x^2 + 3xh + h^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}$

$\lim_{h\to0}\frac{3x^2 + 3xh + h^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\; \frac{3x^2}{x^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{3x^2}{2x^{\frac{3}{2}}} \;=\; \frac{3x^{\frac{1}{2}}}{2}$

Therefore: . $f'(x) \;=\;\tfrac{3}{2}x^{\frac{1}{2}}$

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