Find dy/dx of ln(xy)-sqrt(x)=sqrt(y) Please help me understand this.

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Find dy/dx of ln(xy)-sqrt(x)=sqrt(y) Please help me understand this.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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go @Haseeb96 bhhaiya go
ap kar rahi ho help is ki
noo u have misunderstanding ur doing it

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|dw:1437043335188:dw|
:P see u doing it
|dw:1437043606231:dw|
thanks bhai. hum bhi ek time india mein parhai karte te.
\(\large\color{black}{ \displaystyle \ln(xy)-\sqrt{x}=\sqrt{y} }\) simplify. \(\large\color{black}{ \displaystyle \ln(x)-\ln(y)-\sqrt{x}=\sqrt{y} }\)\ differentiate (dy/dx) \(\large\color{black}{ \displaystyle \frac{1}{x}-\frac{y' }{y}-\frac{1}{2\sqrt{x}}=\frac{y'}{2\sqrt{x}} }\) isolate the terms with y' on one side. \(\large\color{black}{ \displaystyle \frac{1}{x}-\frac{1}{2\sqrt{x}}=\frac{y'}{2\sqrt{x}}+\frac{y' }{y} }\) factor out of y' \(\large\color{black}{ \displaystyle \frac{1}{x}-\frac{1}{2\sqrt{x}}=\left(\frac{1}{2\sqrt{x}}+\frac{1}{y} \right)y'}\) common denominators on both sides \(\large\color{black}{ \displaystyle \frac{2}{2x}-\frac{\sqrt{x}}{2x}=\left(\frac{y}{2y\sqrt{x}}+\frac{2\sqrt{x}}{2y\sqrt{x}} \right)y'}\) adding/subtracting \(\large\color{black}{ \displaystyle \frac{2-\sqrt{x}}{2x}=\left(\frac{y+2\sqrt{x}}{2y\sqrt{x}} \right)y'}\) isolateing the y' (entirely) \(\large\color{black}{ \displaystyle {\LARGE \frac{\frac{2-\sqrt{x}}{2x}}{\frac{y+2\sqrt{x}}{2y\sqrt{x}}} }=y'}\) simplifying further \(\large\color{black}{ \displaystyle \frac{(2-\sqrt{x})~2y\sqrt{x} }{2x~(y+2\sqrt{x})}=y'}\) 2's cancel \(\large\color{black}{ \displaystyle \frac{(2-\sqrt{x})~y\sqrt{x} }{x~(y+2\sqrt{x})}=y'}\) You can play around more, but I would leave it as it is right now.
this made alot of sense, I will go back into the book so im 100% on this. thanks:)

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