Question

derivative of y=x(9-x)^1/2?

Answer 1

\[y=x\sqrt{(9-x)}\]?

Answer 2

yes

Answer 3

product rule for this one

Answer 4

this is what I have:
x[-1/x(9-x)^1/2]+(9-x)^1/2(1)

Answer 5

is that right?

Answer 6

if so how to simplify

Answer 7

\[y'=1\times \sqrt{9-x}+x\times \frac{1}{\sqrt{9-x}}\times -1\] and then clean up with some algebra

Answer 8

really not much to clean up, just
\[y'=\sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]

Answer 9

simplifiedmore?

Answer 10

or if you want to put it over one denominator, you can

Answer 11

\[\frac{9-x-x}{\sqrt{9-x}}\]
\[\frac{9-2x}{\sqrt{9-x}}\]

Answer 12

if you want to "simplify" things usually best not to use exponents, but rather radicals

Answer 13

so√(9-x) *x equals 9-2x right

Answer 14

on no

Answer 15

so how did u getthe numerator

Answer 16

\[y'=\sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]

Answer 17

is that part ok?

Answer 18

oh so u subtract

Answer 19

you want to put i to over the same denominator, so you multiply top and bottom of the first term by
\[\sqrt{9-x}\]

Answer 20

\[y'=\sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]
\[y'=\frac{\sqrt{9-x}}{\sqrt{9-x}}\times \sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]

Answer 21

how come wolfram has: -3(x-6)/2(9-x)^1/2 ?