basic derivatives problem

Soroban · #2 $Old$ October 5th, 2007, 12:23 PM

Hello, afeasfaerw23231233!

Quote:

$\lim_{\Delta x\to0}\,\frac{(x+2)\,(x+\Delta x)^2 - x^2(x + \Delta x + 2)}{\Delta x\, \sqrt{x+2}\,\sqrt{x+\Delta x+2}\,\left[\sqrt{x+2}(x+\Delta x) + x\sqrt{x+\Delta + 2}\right]}$

. . $= \;\lim_{\Delta x\to0}\,\frac{x^2+4x + x\!\cdot\!\Delta x + 2\!\cdot\!\Delta x}{\sqrt{x+2}\,\sqrt{x+\Delta x + 2}\,\left[\sqrt{x+2}(x+\Delta x) + x\sqrt{x+\Delta x + 2}\right]}$

I don't see a way to avoid expanding the numerator.
. . It is, after all, the expected procedure.

The numerator is: . $x^3 + 2x\!\cdot\!\Delta x + x\!\cdot\!\Delta x + 2x^2 + 4x\!\cdot\!\Delta x + 2\!\cdot\!\Delta x^2 - x^3 - x^2\!\cdot\!\Delta x - 2x^2$

. . $= \;x^2\!\cdot\!\Delta x + 4x\!\cdot\!\Delta x + x\!\cdot\!\Delta x + 2\!\cdot\!\Delta x^2 \;=\;\Delta x\,\left(x\!\cdot\!\Delta x + 4x + x\!\cdot\!\Delta x + 2\!\cdot\!\Delta x\right)$

Then: . $\frac{\Delta x\,\left(x\!\cdot\!\Delta x + 4x + x\!\cdot\!\Delta x + 2\!\cdot\!\Delta x\right) } {{\Delta x\, \sqrt{x+2}\,\sqrt{x+\Delta x+2}\,\left[\sqrt{x+2}(x+\Delta x) + x\sqrt{x+\Delta + 2}\right]}}$

. . . . $= \;\frac{x^2+4x + x\!\cdot\!\Delta x + 2\!\cdot\!\Delta x}{\sqrt{x+2}\,\sqrt{x+\Delta x + 2}\,\left[\sqrt{x+2}\left(x + \Delta x\right) + x\sqrt{x + \Delta x + 2}\right]}$

afeasfaerw23231233 · #3 $Old$ October 5th, 2007, 12:34 PM

thanks, but it takes me so much time to expand this piece

!!!