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Derivative of a function, then simplifying it

  • Thread starter stripes
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  • #1
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Hi all...not sure if this should be here, perhaps in the precalc section =S but here goes anyways...

1. So the question asks to find the derivative of y = sqrt(arctan x), then to simplify wherever possible...my issue is not with the differentiation, but rather with the simplifying...oh and we also need to use implicit differentiation when doing so, see my attempt at the solution below.



2. No relevant formulae



3. I basically manipulated the eqn so it was x = tan (sqrt (y)), and then found dy/dx by implicitly differentiating...then i eventually got to this:

dy/dx = (2sqrt (y))/(sec^2 (sqrt(y)))

AND we have y explicitly, so we can substitute y = sqrt(arctan)...so I do that, and I get...

dy/dx = (2sqrt (sqrt(arctan)))/(sec^2 (sqrt(sqrt(arctan))))

which is the same as

dy/dx = (2qrrt(arctan))/(sec^2 (qrrt(arctan))) (qrrt = quartic root of...i'm not sure if that's legit or not...)

so my final answer was dy/dx = (2qrrt(arctan))/(sec^2 (qrrt(arctan))) but the question says simplify, and after an hour of staring at this, i can't seem to figure out what more to simplify?!

Am i missing something? Can i simplify further in this case?

Thanks so much for your help in advance!
 
Last edited:

Answers and Replies

  • #2
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Hey stripes, I have a question - do you have to differentiate implicitly? If you have to find the derivative of

[tex]y=\sqrt{arctan(x)}[/tex],

isn't the equation already in explicit form? (It's been a few years since calculus, so I wasn't sure). In which case, you could do the chain rule:

[tex]\frac{dy}{dx} = \frac{1}{2}(arctan(x))^{-\frac{1}{2}}\left(\frac{1}{1+x^2}\right)[/tex]

Hope this helps. Otherwise, feel free to disregard. :)
 
  • #3
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yes, i failed to mention that in the question lol...we're supposed to use implicit differentiation.

Thanks for the help tho!
 
  • #4
Mentallic
Homework Helper
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stripes, I'm just going to laugh at you and leave it at that... hehe just kidding :wink:

Ok, you see, your problem isn't in the calculus like you said. It's in an even tougher part!

[tex]y=\sqrt{arctan(x)}[/tex]

[tex]y^2=arctan(x)[/tex]

[tex]x=tan(y^2)[/tex]

:tongue:
 
  • #5
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Oh my gosh thank you!!!
 
  • #6
Mentallic
Homework Helper
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No problem :smile:

Those simple mistakes are what frustrates me to no end.
 
  • #7
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yup, and they're always the ones that keep me up all night!

Thanks again!!!!
 
  • #8
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arghh...okay so now i get dy/dx = 1/(sec^2((arctan x))(2sqrt(arctan x))

which simplifies into (cos^2(arctan x))/(2sqrt (arctan x))

so i'm sorta back at the same question...does this simplify further?

Oh, and i don't even think that's dy/dx...it's not equal to what bluskies calculated via explicit differentiation...

what am i doing wrong??
 
  • #9
33,154
4,838
I get the same as what you got for dy/dx. You might get some simplification (or at least in a different form) by evaluating cos^2(arctan x).

Draw a right triangle with opp. side equal to x and adjacent side equal to 1. The hypotenuse is sqrt(1 + x^2).
Let u = arctan x, so tan u = x/1. Using this triangle, you can evaluate cos(u) = cos(arctan x) and hence, cos^2(arctan x).
 

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