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anonymous
 one year ago
Find the derivative
anonymous
 one year ago
Find the derivative

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x)=\sqrt{\left( \frac{ x }{ x^2+3 } \right)^{\arcsin(x^3)}+1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What I did was natural log both sides, but then I get stuck as I can't proceed with the logarithmic differentiation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln(y)=\frac{ 1 }{ 2 }\ln \left( \left( \frac{ x }{ x^2+3 } \right)^{\arcsin(x^3)} +1\right)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Hmm this might seem strange, but I would recommend isolating the big chunk of x stuff before logging, because ya, that +1 is causing a problem.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5\[\large\rm y^21=\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wolfram wont even use this method.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh okay so you're manipulating the inside. So what?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5I'm just thinking that logging from THAT point might be easier:\[\large\rm \ln(y^21)=\arcsin(x^3)\cdot \ln\left(\frac{x}{x^2+3}\right)\]Hmm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How is that equivalent?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like how is what I had before and what you just got equivalent?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Did you not see the steps I took? :o I can explain them again maybe :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now I perfectly get what you did but how is what I had and what you had equivalent?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Hmm that question is confusing... All I did was algebraic manipulation, that shouldn't change anything.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So ln(y^2) = ln(y^21)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is what you wrote though?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5\[\large\rm \ln(y)=\frac{1}{2}\ln \left(\left( \frac{x}{x^2+3}\right)^{\arcsin(x^3)} +1\right)\qquad \text{your equation}\]\[\large\rm \ln(y^21)= \ln\left[\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}\right]\qquad \text{my equation}\]You see that the right side is missing the +1, yes?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do. I suppose I am struggling to see the equivalence because I'm used to seeing it without logs.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Well, if you ignore the logs for a sec, you understand how I went from here,\[\large\rm y=\sqrt{\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}+1}\]to here?\[\large\rm y^21=\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh then you just logged both sides. Gotcha.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooooo. K I get this now.
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