Question

show that for the relation (sq.root of x/y) + (sq.root of y/x) = 10 that dy/dx=y/x HELP!

Answer 1

implicitly derive it im thinking

Answer 2

sqrt(x)/y or is it sqrt(x/y) ??

Answer 3

sqty(x/y)

Answer 4

how well do you understand implicit derivatives?

Answer 5

fairly well at least on a basic level.. this is just a lot harder than any i've done/seen

Answer 6

well, lets see if i can step thru it then

Answer 7

\[\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=10\]
\[\frac{d\ \left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=10\right)}{dx}\]
\[\frac{d\sqrt{\frac{x}{y}}}{dx}+\frac{d\sqrt{\frac{y}{x}}}{dx}=\frac{d(10)}{dx}\]
\[\frac{1}{2\sqrt{\frac{x}{y}}}*\frac{d(\frac{x}{y})}{dx}+\frac{1}{2\sqrt{\frac{y}{x}}}*\frac{d(\frac{y}{x})}{dx}=0\]
so far by the chain rule

Answer 8

\[\frac{d(\frac{x}{y})}{dx}=\frac{yx' - y'x}{y^2}\]
\[\frac{d(\frac{y}{x})}{dx}=\frac{xy' - x'y}{x^2}\]

Answer 9

putting it all together:

\[\frac{1}{2\sqrt{\frac{x}{y}}}*\frac{y-y'x}{y^2}+\frac{1}{2\sqrt{\frac{y}{x}}}*\frac{xy'-y}{x^2}=0\]
\[\frac{y-y'x}{2y^2\sqrt{\frac{x}{y}}}+\frac{xy'-y}{2x^2\sqrt{\frac{y}{x}}}=0\]
\[\frac{y-y'x}{2y^2\sqrt{\frac{x}{y}}}*\frac{x^2\sqrt{\frac{y}{x}}}{x^2\sqrt{\frac{y}{x}}}+\frac{xy'-y}{2x^2\sqrt{\frac{y}{x}}}*\frac{y^2\sqrt{\frac{x}{y}}}{y^2\sqrt{\frac{x}{y}}}=0\]
\[\frac{yx^2\sqrt{\frac{y}{x}}-y'x^3\sqrt{\frac{y}{x}}+xy^2\sqrt{\frac{x}{y}}y'-y^3\sqrt{\frac{x}{y}}}{2x^2y^2\sqrt{\frac{xy}{xy}}}=0\]
\[\frac{y'(xy^2\sqrt{\frac{x}{y}}-x^3\sqrt{\frac{y}{x}})+yx^2\sqrt{\frac{y}{x}}-y^3\sqrt{\frac{x}{y}}}{2x^2y^2}=0\]

Answer 10

\[y'*\frac{xy^2\sqrt{\frac{x}{y}}-x^3\sqrt{\frac{y}{x}}}{2x^2y^2}+\frac{yx^2\sqrt{\frac{y}{x}}-y^3\sqrt{\frac{x}{y}}}{2x^2y^2}=0\]
\[y'*\frac{xy^2\sqrt{\frac{x}{y}}-x^3\sqrt{\frac{y}{x}}}{2x^2y^2}=-\frac{yx^2\sqrt{\frac{y}{x}}-y^3\sqrt{\frac{x}{y}}}{2x^2y^2}\]
\[y'=-\frac{2x^2y^2}{xy^2\sqrt{\frac{x}{y}}-x^3\sqrt{\frac{y}{x}}}*\frac{yx^2\sqrt{\frac{y}{x}}-y^3\sqrt{\frac{x}{y}}}{2x^2y^2}\]
\[y'=-\frac{yx^2\sqrt{\frac{y}{x}}-y^3\sqrt{\frac{x}{y}}}{xy^2\sqrt{\frac{x}{y}}-x^3\sqrt{\frac{y}{x}}}\]
\[y'=-\frac{-y\left(-x^2\sqrt{\frac{y}{x}}+y^2\sqrt{\frac{x}{y}}\right)}{x\left(y^2\sqrt{\frac{x}{y}}-x^2\sqrt{\frac{y}{x}}\right)}\]
notice that that (...) are equal to each other so they cancel to leave us with:
\[y'=-\frac{-y}{x}=\frac{y}{x}\]

Answer 11

its just a mess of algebra really; but it works out in the end

Answer 12

wow thanks so much!

Answer 13

youre welcome; if i hadnt needed the practice, i never would have suffered myself thru that lol