1. ## 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)]

2. 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?

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

4. Can you just explain how you have done that please?

5. 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.

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

7. 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?

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