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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- August 26th 2006, 07:40 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) - August 26th 2006, 07:43 PMThePerfectHackerQuote:

Originally Posted by**startingover**

Thus,

Multiply,

Thus,

- August 26th 2006, 08:03 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) - August 26th 2006, 08:13 PMThePerfectHackerQuote:

Originally Posted by**startingover**

- August 26th 2006, 08:24 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 - August 27th 2006, 12:44 AMGlaysherQuote:

Originally Posted by**startingover**

Have and

- August 27th 2006, 12:45 AMCaptainBlackQuote:

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 - August 27th 2006, 12:47 AMGlaysherQuote:

Originally Posted by**CaptainBlack**

- August 27th 2006, 12:48 AMCaptainBlackQuote:

Originally Posted by**Glaysher**

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

RonL :)