Differentiation - stuck (1 Viewer)

[redorange]

Ok im stuck on differentiating and finding x when A'=0 ...

A = x√2500-x²

A' = ??

.....

I'm using product rule and stuck on..

x = 2500-x²

Could u guys solve it and tell me where i went wrong? thanks.

 

[Mountain.Dew]

A = x√2500-x²

okay, let u = x, u' = 1

v = √(2500-x²) so v' = 1/2 * (2500-x²)^-1/2*(-2x)

so v' = - x / √(2500-x²), by application of the chain rule.

then, simply do A' = uv' + vu' and ur done!


 

[SoulSearcher]

What you have to remember here is the chain rule, where the derivative of √2500-x² is [ 1/2√(2500-x²) ] * -2x. So the answer would be

A = x√(2500-x²)
A' = 1 * √(2500-x²) + (x * -2x) / 2 √(2500-x²)
A' = 5000 - 2x² - 2x² / 2 √(2500-x²)
A' = { 5000 - 4x² } / 2 √(2500-x²)
A' = { 2500 - 2x² } / √(2500-x²)

 

[redorange]

Ahh i see where i went wrong.. thanks guys