# first princple

• Jan 22nd 2010, 10:59 AM
crystal91
first princple
Hi,can someone help me in finding the limit for (5x)^3/2 using first principle.
((5x+h)^3/2-(5x)^3/2)/h
This is all i got and then I am stuck on it.
Help would be greatly appreciated.
Thanx
One more
(3x+5)^3/2
• Jan 22nd 2010, 12:15 PM
Wilmer
• Jan 22nd 2010, 02:02 PM
Soroban
Hello, crystal91!

This is a killer!

Quote:

Differentiate $f(x)\,=\,(5x)^{\frac{3}{2}}$ using first principles.
Note that: . $f(x) \;=\;(5x)^{\frac{3}{2}} \;=\;5^{\frac{3}{2}}\cdot x^{\frac{3}{2}} \;=\; 5\sqrt{5}\,x^{\frac{3}{2}}$

We will differentiate $f(x) \,=\,x^{\frac{3}{2}}$ . . . and multiply by $5\sqrt{5}$ at the very end.

[1] Find $f(x+h) - f(x)$

. . $f(x+h) - f(x) \;\;=\;\;(x+h)^{\frac{3}{2}} - x^{\frac{3}{2}} \;\;=\;\;\left(\sqrt{x+h}\right)^3 - \left(\sqrt{x}\right)^3$ . . . a difference of cubes

. . Factor:. . $\left(\sqrt{x+h} - \sqrt{x}\right)\,\left(x+h + \sqrt{x}\,\sqrt{x+h} + x\right)$ . $=\;\;\left(\sqrt{x+h} - \sqrt{x}\right)\,\left(2x + h + \sqrt{x(x+h)}\right)$

. . Multiply by $\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\!:\quad \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\cdot\frac{(\sqrt{x+h} - \sqrt{x})\,(2x + h + \sqrt{x(x+h)}}{1}$

. . and we have: . $\frac{[(x+h)-x]\,[2x + h + \sqrt{x(x+h)}]}{\sqrt{x+h} + \sqrt{x}} \;=\;\frac{h\,(2x+h + \sqrt{x(x+h)})}{\sqrt{x+h} + \sqrt{x}}$

[2] Divide by $h\!:\quad \frac{2x+h + \sqrt{x(x+h)}}{\sqrt{x+h} + \sqrt{x}}$

[3] Take the limit: . $\lim_{h\to0}\left[\frac{2x+h+\sqrt{x(x+h)}}{\sqrt{x+h} + \sqrt{x}}\right] \;=\;\frac{2x+0+\sqrt{x(x+0)}}{\sqrt{x+0} + \sqrt{x}}$

. . . . . . . . . . . . $=\;\frac{2x + \sqrt{x^2}}{\sqrt{x} + \sqrt{x}} \;\;=\;\;\frac{2x + x}{2\sqrt{x}} \;\;=\;\;\frac{3x}{2\sqrt{x} }\;\;=\;\;\frac{3}{2}\sqrt{x}$

Multiply by $5\sqrt{5}\!:\;\;f'(x) \;=\;\frac{15\sqrt{5}}{2}\,\sqrt{x}$