Question

inverse and inverse trignometric functions

find d/dx(arctan^2(sqrt(x))

Answer 1

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

Answer 2

yea dear

Answer 3

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

Answer 4

sorry. let's start with the derivative of (tan-1u)^2. what is the derivative of this little beastie. lol there we go...

Answer 5

u mean?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

yea

Answer 11

so what is the derivative of tan-1u?

Answer 12

you can give me your best guess if you like

Answer 13

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

Answer 14

sorry 1/1+x^2

Answer 15

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

Answer 16

so thats the answer?

Answer 17

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

Answer 18

is that making sense to you?

Answer 19

still there?

Answer 20

ok

Answer 21

so what will be my final answer?

Answer 22

it is above

Answer 23

can u help me with one other questionÉ

Answer 24

if i can, i would be happy to

Answer 25

find \[find d/dx (\sqrt{1-x ^{2}}+1/2\cos ^{-1}(x) and verify whether or \not this is equal \to (2x+1)d/dx \sin ^{-1}(-x)\]

Answer 26

ok. any ideas where to start?

Answer 27

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

Answer 28

hmm i dont have an idea

Answer 29

ok, well, it is a rather large function, but it is a quotient (hint). So where might we start with this lovely little thing?

Answer 30

hmm..no idea lol

Answer 31

dont know anytin about the topic yet

Answer 32

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

Answer 33

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

Answer 34

do you know the product rule?

Answer 35

yes

Answer 36

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

Answer 37

ok

Answer 38

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

Answer 39

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

Answer 40

yes it does

Answer 41

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?

Answer 42

just algebra after you get the derivatives.

Answer 43

so i should get the derivative of each of them separatelyÉ