- anonymous

inverse and inverse trignometric functions
find d/dx(arctan^2(sqrt(x))

- chestercat

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- anonymous

hi krypton.. you still there? I think i might be able to help if you are

- anonymous

yea dear

- anonymous

consider the derivative of arctan^2(u) where u=sqrt x and use the chain rule...

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## More answers

- 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

u mean?

- anonymous

what is the derivative of (f(x))^2?

- anonymous

it would be something like 2(f(x))times f prime of x

- anonymous

in this case, f(u)=tan-1u, where u=sqrt of x. does this make sense?

- anonymous

questions are good here. the more you ask, the more i can help... :)

- anonymous

yea

- anonymous

so what is the derivative of tan-1u?

- anonymous

you can give me your best guess if you like

- anonymous

hmm actually i wasnt in the class when the topic was taught, but i thinkis =1/srt 1-x^2?

- anonymous

sorry 1/1+x^2

- anonymous

that is totally correct. :) now replace the x with u so that we can use the chain rule where u=1/sqrtx.

- anonymous

so thats the answer?

- anonymous

no, we have a power of tan-1. so the answer is 2(tan-1(sqrtx))(1/(1+x))1/2sqrtx

- anonymous

is that making sense to you?

- anonymous

still there?

- anonymous

ok

- anonymous

so what will be my final answer?

- anonymous

it is above

- anonymous

can u help me with one other questionÃ‰

- anonymous

if i can, i would be happy to

- 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

ok. any ideas where to start?

- anonymous

oh... and i didnt get anything beyond the arrow... sorry... but i think i have enough to figure it out. :)

- anonymous

hmm i dont have an idea

- 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

hmm..no idea lol

- anonymous

dont know anytin about the topic yet

- anonymous

it is ok... remember the quotient rule d(f/g)/dx=[(fprime)g-(gprime)f]/g^2?

- anonymous

in this case the numerator is f=sqrt(1-x^2) +1 and g=2cos-1(x)

- anonymous

do you know the product rule?

- anonymous

yes

- anonymous

phew...!! i am so glad. i thought for a sec we were outta options. lol

- anonymous

ok

- anonymous

look at f(x)=sqrt(1-x^2)+1 and g(x)=(2cos-1(x))^-1. does this make sense?

- anonymous

hahaha i can still
remember my calculus 1, this calculus 2

- anonymous

yes it does

- 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

just algebra after you get the derivatives.

- anonymous

so i should get the derivative of each of them separatelyÃ‰

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