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jkristia
 2 years ago
Problem 5A3g, is the answer correct?
jkristia
 2 years ago
Problem 5A3g, is the answer correct?

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jkristia
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not able to get the same nice clean answer as in the solution, and I think the problem is a mistake in the given solution where he get \[\frac{dy}{dx} \frac{x}{\sqrt{1  x^2}} = (1x^2)^{3/2} ??\] it seems like he forgot the \(x\) in the numerator I get \[=\frac{x^2+x+1}{(1x^2)\sqrt{1x^2}}\] What am I missing, where is my mistake ?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0First change the equation to \[\frac{ dy }{ dx}x(xx ^{2})^{\frac{ 1 }{ 2 }}\] Use the product rule you get \[(1)(1x ^{2})^{\frac{ 1 }{ 2 }}+x(\frac{ 1 }{ 2 })(1x ^{2})^{\frac{3 }{ 2 }}(2x)\] this can be simplified to \[\frac{ 1x ^{2} }{(1x ^{2})^{\frac{3}{2}} }+\frac{ x ^{2} }{(1x ^{2})^{\frac{3}{2}} }\] which equals \[\frac{ 1 }{ (1x ^{2} )^{\frac{ 3 }{ 2 }}}\] or\[(1x ^{2})^{\frac{ 3 }{ 2 }}\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Note that \[(1x ^{2})^{\frac{ 1 }{ 2 }}=\frac{ 1 }{ (1x ^{2})^{\frac{ 1 }{ 2 }} }=\frac{ 1x ^{2} }{ (1x ^{2})^{ \frac{ 3 }{ 2 } }}\]

jkristia
 2 years ago
Best ResponseYou've already chosen the best response.0oh man.... you are right :). I have to go back and check my calculation, but I think my silly mistake was to multiply by (2), not (2x), arghhh... sometime you just go blind staring at a problem.
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