# Chain rule using Leibniz notation.

• Apr 2nd 2010, 12:41 PM
kmjt
Chain rule using Leibniz notation.
Using Leibniz notation, apply the chain rule to determine dy/dx at x=4.

y=u^2+3u, u=square root of x

So I rewrite square root of x as x^1/2

dy/du = 1/2x^-1/2

du/dx = 2u+3

dy/dx = (dy/du)(du/dx)

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

= 1/2x^-1/2 [2(x^1/2)+3] *subbed in u

= 2(x^1/2)+3 / 1/2x^1/2

= 2(x^1/2)+3 / square root of 1/2x

The answer is 7/4, however my answer when I sub in x=4 is 7/1.414213562

What did I do wrong?
• Apr 2nd 2010, 01:02 PM
eddie2042
Quote:

y=u^2+3u, u=square root of x
um.. if you sub in sqrt(x) to the u^2 wouldn't that make it

y = x + 3sqrt(x)?
• Apr 2nd 2010, 01:06 PM
kmjt
= 1/2x^-1/2 (2u+3)

= 1/2x^-1/2 [2(x^1/2)+3]

I subbed in u, which is u=x^1/2, in the u in the 2u+3 part. Wouldn't that make it 2(x^1/2)+3?

Oh I see what you did. Are you supposed to sub in u right away?
• Apr 2nd 2010, 01:07 PM
zzzoak
$\displaystyle \frac{dy}{du}$ = $\displaystyle 2u+3$
$\displaystyle \frac{du}{dx}$ = $\displaystyle \frac{1}{2\sqrt{x}}$

$\displaystyle \frac{dy}{du}\frac{du}{dx}$ = $\displaystyle \frac{2u+3}{2\sqrt{x}}$ = $\displaystyle \frac{2\sqrt{x}+3}{2\sqrt{x}}$ and x=4...
• Apr 2nd 2010, 01:21 PM
kmjt
reworking
• Apr 2nd 2010, 01:34 PM
kmjt
solved it,.. thanks