Question

√xy=x^2+1.. what is dy/dx? , @calculus

Answer 1

\[\sqrt{xy} =x ^{2}+1\]

Answer 2

find the implicit differentiation

Answer 3

yeah I square rooted both sides and did that... I got dy/dx= 4x^2 - y/x but I think it's wrong

Answer 4

i mean squared****

Answer 5

\[\sqrt{xy}=x^2+1\]
\[\frac{1}{2}(xy)^\frac{-1}{2}(xy)'=2x+0\]
\[\frac{1}{2 \sqrt{xy}}(y+xy')=2x\]

Answer 6

\[\frac{y}{2 \sqrt{xy}}+\frac{xy'}{2 \sqrt{xy}}=2x\]

Answer 7

\[\frac{xy'}{2 \sqrt{xy}}=2x-\frac{y}{ 2 \sqrt{xy}}\]

Answer 8

\[y'=\frac{2 \sqrt{xy}}{x}(2x-\frac{y}{2 \sqrt{xy}})\]

Answer 9

\[y'=4 \sqrt{xy}-\frac{y}{x}\]

Answer 10

\[y'=\frac{4x \sqrt{xy}-y}{x}\]

Answer 11

now let's try your way
squaring both sides
\[xy=(x^2+1)^2\]
now differentiate both sides
\[y+xy'=2(x^2+1)(2x+0)\]
\[y+xy'=4x(x^2+1)\]
\[y+xy'=4x \sqrt{xy}\]
\[xy'=4x \sqrt{xy} -y\]
\[y'=\frac{4 x \sqrt{xy} -y}{x}\]
samething :)

Answer 12

\[\text{ remember } \sqrt{xy}=x^2+1\]

Answer 13

so i replaced x^2+1 with (xy)^(1/2) to show we would get the same result

Answer 14

hey money did you delete what you have
did you have
\[y+xy'=4x(x^2+1)\]
\[y+xy'=4x^3+4x\]
\[xy'=4x^3+4x-y\]
\[y'=\frac{4x^3+4x-y}{x}=4x^2+4-\frac{y}{x}\]
this is not wrong

Answer 15

not right you mean?

Answer 16

this is not wrong means this is right

Answer 17

this is one of the ways to write the y'

Answer 18

my book says that to prove that the equation (dy/dx) = \[4\sqrt{x} - (y/x)\]

Answer 19

i have that above

Answer 20

so I'm confused

Answer 21

just divide each term by x

Answer 22

i wrote this above
\[y'=\frac{4 x \sqrt{xy} -y}{x}=4 \sqrt{xy} -\frac{y}{x}\]

Answer 23

but there's an extra y

Answer 24

you mean \[4 \sqrt{xy}-\frac{y}{x} \text{ \right?}\]

Answer 25

i went off money's equation
is that equation written incorrectly?

Answer 26

oh , FAIL I overlooked it. YES That's exactly it! thank you so much myininaya :D