Question

I am finding dy/dx.

This is what I have so far (to be posted in the thread)

Answer 1

\[y=\tan^{-1}(x^2)\]\[\tan(y)=x^2\]\[y'(\sec^2y)=2x\]\[y'=2x/\sec^2y\]\[y'=2x(\cos^2y)\]\[y'=2x(\cos^2(\tan^{-1}x))\]

Answer 2

How do I simplify this further?

Answer 3

rewrite sec^2 as its tangent identity since we know what tan(y) is

Answer 4

what do you mean to write sec^2 as it's tangent identity?

Answer 5

sec in terms of tan?

Answer 6

yes

Answer 7

sec(y)=1/cos(y)=sin(y)/tan(y)

?

Answer 8

(I never liked trig functions always get lost)

Answer 9

recall the pythag setup

sin^2 + cos^2 = 1

divide by cos^2

tan^2 + 1 = sec^2

Answer 10

it is siny times tany

Answer 11

u can directly use this--
\[\frac{ d }{ dx }\tan^{-1} (x)=\frac{ 1 }{ 1+x^2 }\]

Answer 12

then chain rule

Answer 13

2x/sec^2(y)

2x/(tan^2(y) + 1)

2x/((x^2)^2 + 1)

Answer 14

oh, tan^2+1=sec^2

sec y =sqrt{tan^2y+1}

Answer 15

focus lol

we know tan(y) = x^2

therefore: sec^2(y) = tan^2(y) + 1 = x^4 + 1

Answer 16

yes same thing-\[\frac{ d }{ dx }\tan^{-1}(x^2)=\frac{ 1 }{ 1+(x^2)^2 }*d/dx(x^2)\]

Answer 17

and then tan(y)=x^2
tan^2y=(x^2)^2=x^4

so

sec y =sqrt{tan^2y+1}
sec y =sqrt{x^4+1}

Answer 18

it is supposed to be a square root of tat, no?

Answer 19

we have no sec(y) to deal with, just sec^2(y)

Answer 20

we know tan(y) = x^2

therefore: sec^2(y) = tan^2(y) + 1 = x^4 + 1

Answer 21

\[\frac{2x}{sec^2(y)}\]
\[\frac{2x}{tan^2(y)+1}\]
\[\frac{2x}{x^4+1}\]

Answer 22

I see, so

going from \[y'=2x/\sec^2y\]\[y'=2x/(\tan^2y+1)\]\[y'=2x/(x^4+1)\]

Answer 23

yes :)

Answer 24

yeah same:) tnx

Answer 25

and this is it, the answer is there

Answer 26

u can remember derivatives of inverse functions it eases ur calculation

Answer 27

One more thing, how do I re-write
\[y=(\tan^{-1}x)^2\]
Do I say,
\[x=\tan(y^2) ~~~~?\]

Answer 28

with this memory of mine, im lucky if i can remember how to tie my own shoes :)

Answer 29

I am doing proves. I would have used my formula sheet if I was allowed to.
(this is my h/w)

Answer 30

sqrt(y) as long as y is positive

Answer 31

lol ,reminds me of "memento"

Answer 32

or tan(x)=+-sqrt(y) ?

Answer 33

no plus minus?

Answer 34

y = (n^(-1))^(2)

+- sqrt(y) = n^(-1) seems fair

Answer 35

seems fair, but incorrect in this case?

Answer 36

\[y=(\tan^{-1}x)^2\]

\[\pm\sqrt{y}=\tan^{-1}x\]
\[tan(\pm\sqrt{y})=x\]
just work it and see

Answer 37

okay. ty!

Answer 38

tan to the negative one, don't want to put too many parenthesis.

y=(tan^-1 x)^2
+-sqrt(y) = tan-1 x
tan(+- sqrt{y} ) = x
sec^2(+- sqrt{y})=1
sec^2(+- sqrt{y})-1=0
tan^2(+- sqrt{y})

=x

Answer 39

right?

Answer 40

tan(+- sqrt(y))

sec^2(+-sqrt(y)) * (+-1/2sqrt(y)) * y'

Answer 41

ohh, I forgot the chain rule for y.

Answer 42

I think I see what you did

tan^2(+-sqrt(y))
tan [ (+-sqrt(y) ^2 ]

sec^2 [ (+-sqrt(y) ^2 ] because deriv. of tan=sec^2
TIMES
(+- 1/2) * sqrt(y) chain rule for the inner function which is +-sqrt(y) ^2

and TIMES y' as a chain for the derivative of y

Answer 43

\[\frac{\pm2\sqrt{y}}{(tan^2(\pm\sqrt{y})+1)}\]
\[\frac{2~tan^{-1}(x)}{(x^2+1)}\]

Answer 44

I mean 1/2sqrt(y)

Answer 45

the power becomes -1/2

Answer 46

your seeing it ....

Answer 47

I don't want to question you but I think you made a mistake on
deriv. of (+-sqrt(y) ^2

tan [ (+-sqrt(y) ^2 ] =


sec^2 [ (+-sqrt(y) ^2 ] because deriv. of tan=sec^2
TIMES
2/+-sqrt(y) CHAIN for inner function. (but the 2 is on the top)

and TIMES y' (as a chain for y.)

Answer 48

there is no ^2 to start with

Answer 49

ohhh

Answer 50

tan [ (+-sqrt(y) ] = not tan^2[...]

Answer 51

\[y=[tan^{-1}(x)]^2\]
\[\pm \sqrt{y}=tan^{-1}(x)\]
\[tan(\pm \sqrt{y})=x\]
\[sec^2(\pm \sqrt{y})~\frac{\pm1}{2\sqrt y}~y'=x\]

Answer 52

got you

Answer 53

then

( (tan ^2 +-sqrt(y) ) - 1) * 1/(+-2 sqrt y ) * y' =x

(tan ^2(tan-1 x) - 1) * 1/(2(tan-1 x) ) * y' =x

so far correct?

Answer 54

( (tan ^2 +-sqrt(y) ) - 1) * 1/(+-2 sqrt y ) * y' =x
^^
+, not -

Answer 55

( (tan ^2 +-sqrt(y) ) + 1) * 1/(+-2 sqrt y ) * y' =x

Answer 56

( (tan ^2(tan-1 x) ) + 1) * 1/(+-2(tan-1 x) ) * y' =x

Answer 57

( x^2 + 1) * 1/(+-2(tan-1 x) ) * y' =x

Answer 58

\[sec^2(\pm \sqrt{y})~\frac{\pm1}{2\sqrt y}~y'=x\]
\[(\underbrace{tan^2(\pm \sqrt{y})}_{x^2}+1)~\frac{1}{2~\underbrace{(\pm\sqrt y)}_{tan^{-1}(x)}}~y'=x\]

Answer 59

ok...

so after

(x^2+1)(tan^-1 x)y'=x

y' = x
────────
(x^2+1)(tan^-1 x)

Answer 60

a refresh takes care of the <?> parts

Answer 61

\[(x^2+1)~\frac{1}{2~{tan^{-1}(x)}}~y'=1\]
\[(x^2+1)~y'={2~{tan^{-1}(x)}}\]
\[y'=\frac{{2~{tan^{-1}(x)}}}{x^2+1~}\]

Answer 62

true, it is 1/2 sqrt(y)

Answer 63

and in class, the teacher briefly showed how to draw a triangle/angle of tan^-1 x (in a first quadrant), using some Pythagorean theorem... I don't know is it meant to do, and what is it.

Answer 64

its another way to work the setup when you replace y instead of subbing the way we did

Answer 65

for example:

tan(cos^(-1)(x))