Question

use partila derivatis to find dy/dx when y is given implicitly by the equation x^2y-10x=sqrt(xy) + y

Answer 1

\[x^2y-10x=\sqrt{xy}+y\\
{d\over dx}(x^2y)-10{d\over dx}(x)={d\over dx}\sqrt{xy}+{dy\over dx}
\]
use product rule and chain rules for the two of them

Answer 2

lets approach it step by step
what is
\[{d\over dx}(x^2y)=?\]

Answer 3

it is a product.. so you have to use the product rule first
\[{d\over dx}(x^2y)={d\over dx}(x^2)\times y+x^2\times{d\over dx}(y)\]

Answer 4

2xy-10=y/2sqrt(xy) partial of x right?

Answer 5

do not jump.. :)
lets take it slow

Answer 6

did you find the above derivative piece?

Answer 7

2xy+x^2 *d/dx(y)

Answer 8

great.
now, how about
\[{d\over dx}\sqrt{xy}=?\]

Answer 9

y/sqrt(xy)

Answer 10

note.. chain rule first, then product rule

Answer 11

y/2sqrt(xy)

Answer 12

there are two ways to do this.. let me go over the first way
\[\Large{{d\over dx}\sqrt{xy}={d\over dx}(xy)^{1\over2}={1\over2}(xy)^{-1\over2}{d\over dx}(xy)\\
\quad={1\over2(xy)^{1\over2}}\left(1\times y+x\times{dy\over dx}\right)\\
\quad={1\over2\sqrt{xy}}\left(1\times y+x\times{dy\over dx}\right)\\
\quad={y\over2\sqrt{xy}}+{x\over2\sqrt{xy}}{dy\over dx}\\
\quad=\color{red}{{1\over2}\sqrt{y\over x}+{1\over2}\sqrt{x\over y}{dy\over dx}}
}\]

Answer 13

okay. is see the math here.

Answer 14

now we plug em up in the first one.
and solve for \({dy\over dx}\)

Answer 15

before one starts should we set equation to = 0?