## anonymous one year ago Find the derivative

1. anonymous

$f(x)=\sqrt{\left( \frac{ x }{ x^2+3 } \right)^{\arcsin(x^3)}+1}$

2. anonymous

It's a nasty one.

3. zepdrix

yikes 0_o

4. anonymous

What I did was natural log both sides, but then I get stuck as I can't proceed with the logarithmic differentiation.

5. anonymous

$\ln(y)=\frac{ 1 }{ 2 }\ln \left( \left( \frac{ x }{ x^2+3 } \right)^{\arcsin(x^3)} +1\right)$

6. anonymous

Now what?

7. 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.

8. zepdrix

$\large\rm y^2-1=\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}$

9. anonymous

Wolfram wont even use this method.

10. anonymous

Ohh okay so you're manipulating the inside. So what?

11. 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

12. anonymous

How is that equivalent?

13. zepdrix

hmm? :o

14. anonymous

Like how is what I had before and what you just got equivalent?

15. zepdrix

Did you not see the steps I took? :o I can explain them again maybe :D

16. anonymous

Now I perfectly get what you did but how is what I had and what you had equivalent?

17. zepdrix

Hmm that question is confusing... All I did was algebraic manipulation, that shouldn't change anything.

18. anonymous

So ln(y^2) = ln(y^2-1)?

19. zepdrix

no?

20. anonymous

That is what you wrote though?

21. 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?

22. anonymous

I do. I suppose I am struggling to see the equivalence because I'm used to seeing it without logs.

23. 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?$\large\rm y^2-1=\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}$

24. anonymous

Yep I got that.

25. anonymous

Ohh then you just logged both sides. Gotcha.

26. anonymous

ooooo. K I get this now.

27. anonymous

Thanks!

28. zepdrix

>.< heh

29. imqwerty

:)