# Determining dy/du, du/dx, dy/dx

• May 17th 2010, 04:13 PM
oregon88
Determining dy/du, du/dx, dy/dx
$y={\sqrt u} + \frac {1}{\sqrt u}$

I found out what $\frac{dy}{du}$ is and what $\frac{du}{dx}$ and when I do $\frac{dy}{dx}$ I get $\frac {(3x^2)(x^3-1)}{2(x^3)^\frac {3}{2}}$ is this correct? or is it just $\frac {(3x^2)(x^3)}{2(x^3)^\frac {3}{2}}$
• May 17th 2010, 04:38 PM
skeeter
Quote:

Originally Posted by oregon88
$y={\sqrt u} + \frac {1}{\sqrt u}$

I found out what $\frac{dy}{du}$ is and what $\frac{du}{dx}$ and when I do $\frac{dy}{dx}$ I get $\frac {(3x^2)(x^3-1)}{2(x^3)^\frac {3}{2}}$ is this correct? or is it just $\frac {(3x^2)(x^3)}{2(x^3)^\frac {3}{2}}$

how is u related to x ?
• May 17th 2010, 04:51 PM
oregon88
$\frac {dy}{du} = \frac {1}{2}u^\frac{-1}{2}-\frac{1}{2}u^\frac{-3}{2}$

$\frac{du}{dx} = 3x^2$
• May 17th 2010, 05:29 PM
skeeter
Quote:

Originally Posted by oregon88
$\frac {dy}{du} = \frac {1}{2}u^\frac{-1}{2}-\frac{1}{2}u^\frac{-3}{2}$

$\frac{du}{dx} = 3x^2$

ok ... $\frac{du}{dx} = 3x^2$ ... is $u = x^3 + 1000$ ?
• May 17th 2010, 05:41 PM
oregon88
$u=x^3-x$
• May 17th 2010, 05:49 PM
skeeter
$\frac{d}{dx}\left[y = \sqrt{u} + \frac{1}{\sqrt{u}}\right]
$

$\frac{dy}{dx} = \left[\frac{1}{2\sqrt{u}} - \frac{1}{2\sqrt{u^3}}\right] \cdot \frac{du}{dx}$

if $u = x^3 - x$

$\frac{du}{dx} = 3x^2 - 1$

$\frac{dy}{dx} = \left[\frac{1}{2\sqrt{x^3-x}} - \frac{1}{2\sqrt{(x^3-x)^3}}\right] \cdot (3x^2-1)$
• May 17th 2010, 06:03 PM
oregon88
When you simplify that you get $\frac{(3x^2-1)(x^3-x-1)}{2(x^3-x)^\frac{3}{2}}$ but what if $u=x^3$? would you get $\frac {(3x^2)(x^3-1)}{2(x^3)^\frac {3}{2}}$ or $\frac {(3x^2)(x^3)}{2(x^3)^\frac {3}{2}}$?