# Simple differentiation!

• May 21st 2011, 10:54 AM
ishida
Simple differentiation!
I am very out of practice with differentiation as I have not done it for over three years. However, I have just come across this problem and I need to solve it as soon as possible. I have tried reading around about how to do it, but to no avail.

Get dx/dy when

x = ay[1-((y/b)^3)]

• May 21st 2011, 11:08 AM
ishida
This may even be A-Level actually.

Anyway I just used the product rule to break it down into two parts

u=ay v=[1-((y/b)^3)]

differentiating u is easy, but now I am struggling differentiating v. help?
• May 21st 2011, 11:10 AM
$\displaystyle v = 1 - {\frac{y}{ b} }^{ 3}$ $\displaystyle dv = -3{\frac{y}{ b}}^{ 2}$
• May 21st 2011, 11:13 AM
ishida
Can you just explain how you have done that please?
• May 21st 2011, 11:16 AM
Quote:

Originally Posted by ishida
Can you just explain how you have done that please?

If it is with respect to y (which i have assumed), all you do is times the y term by the power, then reduce the power by 1.
• May 21st 2011, 11:20 AM
Is $\displaystyle v = 1 - {\frac{y}{ b} }^{ 3}$ or is $\displaystyle v = 1 - {(\frac{y}{ b}) }^{ 3}$ ??
• May 21st 2011, 11:20 AM
ishida
Ok, makes sense. However, in the solution I have been given there appears to be another term at the end. 1/b. Where has this come from?
• May 21st 2011, 11:21 AM
ishida
Quote:

Originally Posted by adam_leeds
Is $\displaystyle v = 1 - {\frac{y}{ b} }^{ 3}$ or is $\displaystyle v = 1 - {(\frac{y}{ b}) }^{ 3}$ ??

The second of those.
• May 21st 2011, 11:23 AM
$\displaystyle v = 1 - {(\frac{y}{ b}) }^{ 3}$ You times just the right hand side by the power and the differntation of the brackets with respect to y which is $\displaystyle 1/b$ then reduce the power by 1. It's called the chain rule :)