anonymous
  • anonymous
Find dy/dx of ln(xy)-sqrt(x)=sqrt(y) Please help me understand this.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
go @Haseeb96 bhhaiya go
Haseeb96
  • Haseeb96
ap kar rahi ho help is ki
anonymous
  • anonymous
noo u have misunderstanding ur doing it

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Haseeb96
  • Haseeb96
|dw:1437043335188:dw|
anonymous
  • anonymous
:P see u doing it
Haseeb96
  • Haseeb96
|dw:1437043606231:dw|
anonymous
  • anonymous
thanks bhai. hum bhi ek time india mein parhai karte te.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \ln(xy)-\sqrt{x}=\sqrt{y} }\) 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.
anonymous
  • anonymous
this made alot of sense, I will go back into the book so im 100% on this. thanks:)

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