anonymous
  • anonymous
Find the derivative
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[f(x)=\sqrt{\left( \frac{ x }{ x^2+3 } \right)^{\arcsin(x^3)}+1}\]
anonymous
  • anonymous
It's a nasty one.
zepdrix
  • zepdrix
yikes 0_o

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anonymous
  • anonymous
What I did was natural log both sides, but then I get stuck as I can't proceed with the logarithmic differentiation.
anonymous
  • anonymous
\[\ln(y)=\frac{ 1 }{ 2 }\ln \left( \left( \frac{ x }{ x^2+3 } \right)^{\arcsin(x^3)} +1\right)\]
anonymous
  • anonymous
Now what?
zepdrix
  • zepdrix
Hmm 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
  • zepdrix
\[\large\rm y^2-1=\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}\]
anonymous
  • anonymous
Wolfram wont even use this method.
anonymous
  • anonymous
Ohh okay so you're manipulating the inside. So what?
zepdrix
  • zepdrix
I'm just thinking that logging from THAT point might be easier:\[\large\rm \ln(y^2-1)=\arcsin(x^3)\cdot \ln\left(\frac{x}{x^2+3}\right)\]Hmm
anonymous
  • anonymous
How is that equivalent?
zepdrix
  • zepdrix
hmm? :o
anonymous
  • anonymous
Like how is what I had before and what you just got equivalent?
zepdrix
  • zepdrix
Did you not see the steps I took? :o I can explain them again maybe :D
anonymous
  • anonymous
Now I perfectly get what you did but how is what I had and what you had equivalent?
zepdrix
  • zepdrix
Hmm that question is confusing... All I did was algebraic manipulation, that shouldn't change anything.
anonymous
  • anonymous
So ln(y^2) = ln(y^2-1)?
zepdrix
  • zepdrix
no?
anonymous
  • anonymous
That is what you wrote though?
zepdrix
  • zepdrix
\[\large\rm \ln(y)=\frac{1}{2}\ln \left(\left( \frac{x}{x^2+3}\right)^{\arcsin(x^3)} +1\right)\qquad \text{your equation}\]\[\large\rm \ln(y^2-1)= \ln\left[\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}\right]\qquad \text{my equation}\]You see that the right side is missing the +1, yes?
anonymous
  • anonymous
I do. I suppose I am struggling to see the equivalence because I'm used to seeing it without logs.
zepdrix
  • zepdrix
Well, 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?\[\large\rm y^2-1=\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}\]
anonymous
  • anonymous
Yep I got that.
anonymous
  • anonymous
Ohh then you just logged both sides. Gotcha.
anonymous
  • anonymous
ooooo. K I get this now.
anonymous
  • anonymous
Thanks!
zepdrix
  • zepdrix
>.< heh
imqwerty
  • imqwerty
:)

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