Inverse Trigonometry question

[cedar tulip]

I get how dx/dy = 2 · sec^2x · tanx

but I don't understand how sec^2y = 1/(1+y^2). Can't use a triangle because its in terms of y and from my understanding, 1/(1+y^2) is dx/dy tan^-1y???

can someone help explain steps 3 to 4

[worn flare]

In 3rd step we take square on both sides then take Differentiation with respect to dy

Then use chain rule for tan^2 = 2x 1/1+y^2 X tany

[cedar tulip]

how did you get 1/1+y^2 using chain rule?

[worn flare]

1/1+y^2 is derivative of tan^-1

[cedar tulip]

yes so where did tan^-1 come from

[mighty heron]

your's is correct
the derivative of tan(y) is sec²(y)
so dx/dy = 2×sec²(y)×tan(y)

[cedar tulip]

so the solution is incorrect?

[mighty heron]

yea

[cedar tulip]

can it be simplified further?

here is the rest of the solution

[mighty heron]

dy/dx = 1/(2×sec²y×tan y) = cos²y/(2 tan y)

or you can leave it as it is

[cedar tulip]

because the answer cannot be left in terms of y

[mighty heron]

then substitute y with arctan(sqrt(x))

[cedar tulip]

right

do i leave sec^2(tan^-1(sqrt(x)) as it is or is there a way to simplify it further

[mighty heron]

leave it as it is

[cedar tulip]

ok thank you

[mighty heron]

urwelcome

[cedar tulip]

hey i did some messing around and it turns out that sec^2(tan^-1(x)) = 1 + x^2

but i don't understand why

hang on i worked it out

sec^2(tan^-1(x)) can be rewritten as 1 + tan^2(tan^-1(x)) = 1 + x^2

[latent blaze]

there is also a geometric explanation

based on the unit circle

but you most likely don't need to know this until you learn trig substitution

[cedar tulip]

is that highschool maths?

[latent blaze]

depends what modules you do

might be in calc bc

[cedar tulip]

i've looked through my formula sheet and I can't find any thing related to 'trig substitution' so its probably not in our syllabus

that looks difficult so i'm thankful I don't have to learn it