Question

For the curve sqrt(xy)-x+y = 1, find the value of dy/dx at the point (9,4)

Answer 1

differentiate
\[\frac{y + xy'}{2 \sqrt{xy}} -1 + y' = 0\]
solve for y'

Answer 2

but where did the 2sqrt(xy) come from?

Answer 3

it comes from derivative of sqrt(xy)

you use chain rule
\[\frac{d}{dx} \sqrt{u} = \frac{1}{2 \sqrt{u}} \frac{du}{dx}\]

Answer 4

ok so for every square root, one must follow that correct?

Answer 5

yes it comes from power rule
\[\sqrt{x} = x^{1/2} \rightarrow (x^{1/2})' = (1/2) x^{-1/2}\]

Answer 6

oooh i see...

Answer 7

now on the top part, the y + xy', is what the derivative of?

Answer 8

the "xy" inside the sqrt

it comes from product rule

Answer 9

oh i get it! now with that do we just plug in (9,4) into the x and y?

Answer 10

not yet, first you need to isolate y', then plug in the point to get value for y'

Answer 11

or i guess it doesn't matter, you can plug in the values first , then solve for y'

Answer 12

allrighty!

Answer 13

so does 3y'/4 = -3?

Answer 14

i keep getting y' = 4/3, but that is not an answer choice, am i doing something wrong?

Answer 15

thats not what im getting

Answer 16

\[\frac{4 +9y'}{12} + y' = 1\]
\[4 +9y' +12y' = 12\]
\[21y' = 8\]

...

Answer 17

wait, why did you make it =12 and have 12y'?

Answer 18

oh to get rid of the fraction, i multiplied everything by the denominator

Answer 19

whoah! i didn't know you could do that! sudden realization of greater truth! then y' = 8/21

Answer 20

haha oh yeah, remember to do that it will make the algebra a bit easier :)

Answer 21

allrighty! thank you sooo much!!

Answer 22

yw