Question

what is the derivative of:

Answer 1

\[A(x)=(2x)(\sqrt{r ^{2}-x ^{2}})\]

Answer 2

where \(r\) is constant I guess?

Answer 3

This is a mix of the product rule and the chain rule then.

Answer 4

how?

Answer 5

Product rule is

\[(uv)'= u'v + v'u\]

Answer 6

For the radical expression you need to apply the chain rule.

Answer 7

okay..thanks! :)

Answer 8

welcome, gives it a try, we can afterwards check your answer together or you can enter it in wolframalpha.

Answer 9

okay..later! :)

Answer 10

is the answer:
\[A'=(2r ^{2}+2xr) \div (\sqrt{r ^{2}-x ^{2}})\] ?

Answer 11

i'm solving for a calculus problem..
so i must equate that derivative to zero..and solve for x...
how?

Answer 12

i'm confused

Answer 13

how'd you get that? @Spacelimbus

Answer 14

let me try that again step by step before i make silly mistakes.

Answer 15

\[\Large A'= 2\sqrt{r^2-x^2} - \frac{2x^2}{\sqrt{r^2-x^2}} \]

Answer 16

which is equal to \[ \Large A'= \frac{2(r^2-2x^2)}{\sqrt{r^2-x^2}} \]