simple derivation with radicals

[rubing]

This is a pretty simple equation that has radicals, which I need to solve. I forgot how to solve these...can someone please help? thanks!!


find g'(x) for g(x) = x√x


so far I have:


∆y + y = (x + ∆x) √(x + ∆x)


∆y = (x + ∆x) √(x + ∆x) - x√x


∆y / ∆x = [(x + ∆x) √(x + ∆x) - x√x] / ∆x

 

[soroban]

Hello, rubing!

\(\displaystyle \text{Find }g'(x)\text{ for: }\,g(x) \,=\,x\sqrt{x}\)

We have: .\(\displaystyle y \:=\:x^{\frac{3}{2}}\)

\(\displaystyle y + \Delta y \;=\;(x+\Delta x)^{\frac{3}{2}} - x^{\frac{3}{2}}\quad\hdots\;\text{ Now rationalize the numerator}\)

. . . . . . .\(\displaystyle =\;\dfrac{(x+\Delta x)^{\frac{3}{2}} - x^{\frac{x}{3}}}{1} \cdot \dfrac{(x+\Delta x)^{\frac{3}{2}} + x^{\frac{3}{2}}}{(x+\Delta x)^{\frac{3}{2}} + x^{\frac{3}{2}}}\)

. . . . . . .\(\displaystyle =\;\dfrac{(x+\Delta x)^3 - x^3}{(x+\Delta x)^{\frac{3}{2}} + x^{\frac{3}{2}}} \)

. . . . . . .\(\displaystyle =\; \dfrac{x^3 + 3x^2(\Delta x) + 3x(\Delta x)^2 + (\Delta x)^3 - x^3}{(x+\Delta x)^{\frac{3}{2}} + x^{\frac{3}{2}}}\)

. . . . . . .\(\displaystyle =\;\dfrac{3x^2(\Delta x) + 3x(\Delta x)^2 + (\Delta x)^3}{(x+\Delta x)^{\frac{3}{2}} + x^{\frac{3}{2}}} \)

. . . . . . .\(\displaystyle \;=\;\dfrac{(\Delta x)\big[3x^2 + 3x(\Delta x) + (\Delta x)^2\big]}{(x+\Delta x)^{\frac{3}{2}} + x^{\frac{3}{2}}} \)


\(\displaystyle \dfrac{y+\Delta y}{\Delta x} \;=\;\dfrac{3x^2 + 3x(\Delta x) + (\Delta x)^2}{(x+\Delta x)^{\frac{3}{2}} + x^{\frac{3}{2}}} \)


Got it?