Simple differentiation!

**ishida** · May 21st 2011, 10:54 AM

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)]

Please help!

**ishida** · May 21st 2011, 11:08 AM

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?

**adam_leeds** · May 21st 2011, 11:10 AM

$v = 1 - {\frac{y}{ b} }^{ 3}$ $dv = -3{\frac{y}{ b}}^{ 2}$

**ishida** · May 21st 2011, 11:13 AM

Can you just explain how you have done that please?

**adam_leeds** · May 21st 2011, 11:16 AM

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.

**adam_leeds** · May 21st 2011, 11:20 AM

Is $v = 1 - {\frac{y}{ b} }^{ 3}$ or is $v = 1 - {(\frac{y}{ b}) }^{ 3}$ ??

**ishida** · May 21st 2011, 11:20 AM

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?

**ishida** · May 21st 2011, 11:21 AM

Originally Posted by adam_leeds

Is $v = 1 - {\frac{y}{ b} }^{ 3}$ or is $v = 1 - {(\frac{y}{ b}) }^{ 3}$ ??

The second of those.

**adam_leeds** · May 21st 2011, 11:23 AM

$v = 1 - {(\frac{y}{ b}) }^{ 3}$ You times just the right hand side by the power and the differntation of the brackets with respect to y which is $1/b$ then reduce the power by 1. It's called the chain rule

**ishida** · May 21st 2011, 11:24 AM

Makes sense. Brilliant. Thank you for your help

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