Derivative/chain rule

**daigo** · July 13th 2012, 05:37 PM

$f(x) = \frac{1}{x\sqrt{5-2x}}$

$ f'(x) = (x\sqrt{5-2x})^{-1} \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (5 - 2x)') \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (- 2))) \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}) \\ = -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)} $

Have I done anything wrong so far?

**daigo** · July 13th 2012, 06:04 PM

After I simplify, I get this:

$= \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)} \\ \\ \\ = \frac{1}{x^{2}(5 - 2x)^{\frac{1}{2}}} - \frac{1}{x(5-2x)^{\frac{3}{2}}}$

But the solution says I'm wrong...what have I done?

**daigo** · July 13th 2012, 06:20 PM

Oh wait, I see it now...the very first negative sign at the very beginning...I completely missed that this entire time and I couldn't see it...wow

**Reckoner** · July 13th 2012, 06:20 PM

Originally Posted by daigo

$= -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)}$

After I simplify, I get this:

$= \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)}$

You forgot the negative sign that was in front. That should be

$f'(x) = -\frac{(5-2x)^{1/2}}{x^2(5-2x)} + \frac{x(5-2x)^{-1/2}}{x^2(5-2x)}$

$=\frac1{x(5-2x)^{3/2}} - \frac1{x^2(5-2x)^{1/2}}$

This is the correct derivative. Naturally, your textbook might present it in a different form.

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