anonymous
  • anonymous
inverse and inverse trignometric functions find d/dx(arctan^2(sqrt(x))
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
hi krypton.. you still there? I think i might be able to help if you are
anonymous
  • anonymous
yea dear
anonymous
  • anonymous
consider the derivative of arctan^2(u) where u=sqrt x and use the chain rule...

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anonymous
  • anonymous
sorry. let's start with the derivative of (tan-1u)^2. what is the derivative of this little beastie. lol there we go...
anonymous
  • anonymous
u mean?
anonymous
  • anonymous
what is the derivative of (f(x))^2?
anonymous
  • anonymous
it would be something like 2(f(x))times f prime of x
anonymous
  • anonymous
in this case, f(u)=tan-1u, where u=sqrt of x. does this make sense?
anonymous
  • anonymous
questions are good here. the more you ask, the more i can help... :)
anonymous
  • anonymous
yea
anonymous
  • anonymous
so what is the derivative of tan-1u?
anonymous
  • anonymous
you can give me your best guess if you like
anonymous
  • anonymous
hmm actually i wasnt in the class when the topic was taught, but i thinkis =1/srt 1-x^2?
anonymous
  • anonymous
sorry 1/1+x^2
anonymous
  • anonymous
that is totally correct. :) now replace the x with u so that we can use the chain rule where u=1/sqrtx.
anonymous
  • anonymous
so thats the answer?
anonymous
  • anonymous
no, we have a power of tan-1. so the answer is 2(tan-1(sqrtx))(1/(1+x))1/2sqrtx
anonymous
  • anonymous
is that making sense to you?
anonymous
  • anonymous
still there?
anonymous
  • anonymous
ok
anonymous
  • anonymous
so what will be my final answer?
anonymous
  • anonymous
it is above
anonymous
  • anonymous
can u help me with one other question√Č
anonymous
  • anonymous
if i can, i would be happy to
anonymous
  • anonymous
find \[find d/dx (\sqrt{1-x ^{2}}+1/2\cos ^{-1}(x) and verify whether or \not this is equal \to (2x+1)d/dx \sin ^{-1}(-x)\]
anonymous
  • anonymous
ok. any ideas where to start?
anonymous
  • anonymous
oh... and i didnt get anything beyond the arrow... sorry... but i think i have enough to figure it out. :)
anonymous
  • anonymous
hmm i dont have an idea
anonymous
  • anonymous
ok, well, it is a rather large function, but it is a quotient (hint). So where might we start with this lovely little thing?
anonymous
  • anonymous
hmm..no idea lol
anonymous
  • anonymous
dont know anytin about the topic yet
anonymous
  • anonymous
it is ok... remember the quotient rule d(f/g)/dx=[(fprime)g-(gprime)f]/g^2?
anonymous
  • anonymous
in this case the numerator is f=sqrt(1-x^2) +1 and g=2cos-1(x)
anonymous
  • anonymous
do you know the product rule?
anonymous
  • anonymous
yes
anonymous
  • anonymous
phew...!! i am so glad. i thought for a sec we were outta options. lol
anonymous
  • anonymous
ok
anonymous
  • anonymous
look at f(x)=sqrt(1-x^2)+1 and g(x)=(2cos-1(x))^-1. does this make sense?
anonymous
  • anonymous
hahaha i can still remember my calculus 1, this calculus 2
anonymous
  • anonymous
yes it does
anonymous
  • anonymous
glad to hear it... calc I is totally needed here... good, it makes sense. now we need fprime and gprime and i think you are pretty much homefree. so what is fprime and what is gprime?
anonymous
  • anonymous
just algebra after you get the derivatives.
anonymous
  • anonymous
so i should get the derivative of each of them separately√Č

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