# Thread: chain rule and derivatives

1. ## chain rule and derivatives

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)

2. Originally Posted by startingover

y = u^2 + u - 2
u= x^1/2
$\displaystyle y'=2u+1$
$\displaystyle u=1/(2\sqrt{x})$
Thus,
$\displaystyle 2(1/2(\sqrt{x}))+1=1/\sqrt{x}+1$
Multiply,
$\displaystyle 1/(2\sqrt{x})(1/\sqrt{x}+1)$
Thus,
$\displaystyle 1/(2x)+1/(2\sqrt{x})$

3. ## Hmmmm....

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)

4. Originally Posted by startingover
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)
Seems you know this stuff better than I do. Previous post has been corrected.

5. ## ummmm...

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

6. Originally Posted by startingover

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)
If $\displaystyle u = \frac{1}{x}$ then $\displaystyle u \ne x^{\frac{1}{2}}$

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

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

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

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

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

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

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

$\displaystyle = -\frac{2}{x^3} -\frac{1}{x^2} = -\frac{2 + x}{x^3}$

7. Originally Posted by startingover

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)
Using the ' notation for derivative here is confusing as it means d/dx in one
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

8. Originally Posted by CaptainBlack
Using the ' notation for derivative here is confusing as it means d/dx in one
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
You missed out the minus from du/dx

9. Originally Posted by Glaysher
You missed out the minus from du/dx
Noticed it and went back and put it in before seeing your note

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

RonL