Question

Find y '' by implicit diferentiation
sqrt of x + sqrt of y = 1

Answer 1

\[\sqrt{x}+\sqrt{y} = 1\]

Answer 2

\[\frac{1}{2} x^{-\frac{1}{2}} + \frac{1}{2} y^{- \frac{1}{2}} \frac{dy}{dx} = 0 \]

Answer 3

\[\frac{1}{2 \sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} =0 \]

Answer 4

\[\frac{dy}{dx} = - \sqrt{ \frac{y}{x}}\]

Answer 5

y do u have to multiply that by dy/dx

Answer 6

\[\frac{d}{dx} [ \frac{dy}{dx} ] = - \frac{d}{dx} [ (\frac{y}{x})^{\frac{1}{2}}] \]

Answer 7

because it's a chain rule.

Answer 8

were u isolating dy/dx to get -sqrt y/x?

Answer 9

Yeah

Answer 10

can u plz show the steps to how u got taht....

Answer 11

i get confused if steps r skipped

Answer 12

@kira u have dy/dx is positive whereas elec has it negaitve

Answer 13

Oops... I left out -ve sign when I was typing for the screenshot... Gomen...

Answer 14

im sorry but the math looks realllllly confusing...esp the last step where u were solving d^2y/dx^2

Answer 15

do u mind explaing wat u were tryin to do

Answer 16

Here's an errata, hope it doesn't contain any mistakes. I'm using quotient rule