Compute the derivative dy/dx and simplify your answer.

y = u^2 + u - 2

u = 1/x or x^1/2

y' = 2u + 1

u' = 1/2(x)^-1/2 = 1/2(square root x)

2(1/x) + 1 *1/2(square root x)

2/x + 1 * 1/2(square root x)

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- Aug 26th 2006, 06:40 PMstartingoverchain rule and derivatives
Compute the derivative dy/dx and simplify your answer.

y = u^2 + u - 2

u = 1/x or x^1/2

y' = 2u + 1

u' = 1/2(x)^-1/2 = 1/2(square root x)

2(1/x) + 1 *1/2(square root x)

2/x + 1 * 1/2(square root x) - Aug 26th 2006, 06:43 PMThePerfectHackerQuote:

Originally Posted by**startingover**

$\displaystyle u=1/(2\sqrt{x})$

Thus,

$\displaystyle 2(1/2(\sqrt{x}))+1=1/\sqrt{x}+1$

Multiply,

$\displaystyle 1/(2\sqrt{x})(1/\sqrt{x}+1)$

Thus,

$\displaystyle 1/(2x)+1/(2\sqrt{x})$ - Aug 26th 2006, 07:03 PMstartingoverHmmmm....
I'm sorry, but I'm a bit confused.

I thought you substituted the orginal u expression (1/x) into the derivative of y (2u + 1), then multiplied by the derivative of u (1/2squ.rt.x).

2[1/x] + 1 (1/2squ.rt.x) - Aug 26th 2006, 07:13 PMThePerfectHackerQuote:

Originally Posted by**startingover**

- Aug 26th 2006, 07:24 PMstartingoverummmm...
Not trying to be a pain - honest...

In the beginning equation, why did you substitute 1/2square rt x in for the u expression? Shouldn't it be the orginal u, not the derivative of u?

y = u^2 + u - 2

y' = 2u + 1

u = 1/x

u' = 1/2squarert x

2 (1/x) + 1 * 1/2squarert x - Aug 26th 2006, 11:44 PMGlaysherQuote:

Originally Posted by**startingover**

$\displaystyle u = x^{-1}$

$\displaystyle u' = -x^{-2}$

Have $\displaystyle \frac{dy}{du} = 2u + 1$ and $\displaystyle \frac{du}{dx} = -x^{-2}$

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

$\displaystyle = (2u + 1)(-x^{-2})$

$\displaystyle = (2x^{-1} + 1)(-x^{-2})$

$\displaystyle = -2x^{-3} - x^{-2}$

$\displaystyle = -\frac{2}{x^3} -\frac{1}{x^2} = -\frac{2 + x}{x^3}$ - Aug 26th 2006, 11:45 PMCaptainBlackQuote:

Originally Posted by**startingover**

place and d/dy in another.

The chain rule is:

dy/dx=dy/du du/dx

so:

dy/du=2u+1

and:

du/dx=-1/x^2 (that is when u=1/x)

So:

dy/dx=-(2u+1)/x^2=-(2/x+1)/x^2=-(x+2)/x^3

RonL - Aug 26th 2006, 11:47 PMGlaysherQuote:

Originally Posted by**CaptainBlack**

- Aug 26th 2006, 11:48 PMCaptainBlackQuote:

Originally Posted by**Glaysher**

(peculiar that post does not show time it was edited)

RonL :)