1. ## another differentiation question

if we have the above, then how would the answer change if we changed
d(v/x)/dx to:
1) d(2v/x)/dx
2) d(v^2/x)/dx
3) d((v+3)/x)/dx

thanks

2. few question before we start.

Are you able to derive the original result ? does the quotient rule mean anything to you?

Bobak

3. Originally Posted by Richyie

if we have the above, then how would the answer change if we changed
d(v/x)/dx to:
1) d(2v/x)/dx
2) d(v^2/x)/dx
3) d((v+3)/x)/dx

thanks
Just use this $\displaystyle \frac{D[\frac{u}{v}]}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

4. then would this be right:

d(v^2/x)/dx = ( 2vx(dv/dx) - v^2 ) / x^2

thnx

5. Originally Posted by Richyie
then would this be right:

d(v^2/x)/dx = ( 2vx(dv/dx) - v^2 ) / x^2

thnx
$\displaystyle \frac{D[\frac{v^2}{x}]}{dx}=\frac{x\cdot{2}\cdot{v}\cdot\frac{dv}{dx}-v^2}{x^2}$

6. ## Ok

Originally Posted by Mathstud28
$\displaystyle \frac{D[\frac{v^2}{x}]}{dx}=\frac{x\cdot{2}\cdot{v}\cdot\frac{dv}{dx}-v^2}{x^2}$
Just as a disclaimer...this answer is valid if $\displaystyle v(x)$ is what $\displaystyle v$ implied and not a constant