Inverse Trig Question (1 Viewer)

[star*eyed]

I have to answer the following question:

Find the value of the derivative of i inverse sin (tan x) at x = 0

I get an answer of 1, but i don't know if that is correct, could anyone verify if im right or wrong and if im wrong point me in the right direction???

THANX!

 

[Trev]

That's right.

 

[shinji]

how did u get 1? lol.. care to enlighten us?

 

[pLuvia]

d/dx[arcsin(tanx)]
=sec²x[1/(1-tan²x)]
When x=0
dy/dx=sec²0[1/(1-tan²0)]
=1

 

[Porcia]

d/dx[arcsin(tanx)]
=sec²x[1/(1-tan²x)]
When x=0
dy/dx=sec²0[1/(1-tan²0)]
=1

correct me if im wrong but isnt the inverse of sin derivation: 1/(rt:1-tan(^2)x) and not in fact the one you had displayed? but that seems to give 1/rt:2 as an answer instead so its probably wrong

 

[Riviet]

correct me if im wrong but isnt the inverse of sin derivation: 1/(rt:1-tan(^2)x) and not in fact the one you had displayed? but that seems to give 1/rt:2 as an answer instead so its probably wrong

pLuvia is correct, since in the derivative of sin^-1(tanx), you have to multiply by the derivative of the inside function, tanx, that is:

d/dx(sin^-1[f(x)]) = f'(x)/sqrt(1-[f(x)]²)

 

[Trev]

Don't you have school Riviet?

 

[Riviet]

Nope, I don't think anyone does, since it's the Queen's Birthday long weekend.

 

[Trev]

It is?... lol. Righto.

 

[Porcia]

pLuvia is correct, since in the derivative of sin^-1(tanx), you have to multiply by the derivative of the inside function, tanx, that is:

d/dx(sin^-1[f(x)]) = f'(x)/sqrt(1-[f(x)]²)

which means riviet is in fact wrong... lol... go ahead and try out the math in full - without any recourse from this forum

 

[SoulSearcher]

I think pLuvia just forgot the square root sign in the denominator, but it still works out to be the same answer:
d/dx{arcsin(tan x)} = sec²x[1/rt(1-tan²x)]
= sec²x/rt(sec²x)
When x = 0,
dy/dx = sec²0/rt(sec²0)
= 1/rt(1)
= 1

EDIT: Screwed it up in this post.

 

[Porcia]

still wrong: the correct answer is root two... 1-tan^2x does not equal sec^2x. it equals 2-sec^2.

someone please agree with me otherwise youse all will shake my foundation of mathematical learning... lol!

 

[SoulSearcher]

Ok, let's do this differentiation slowly.
y = sin^-1(tanx)
= sin^-1u, where u = tan x
du/dx = sec²x
dy/du = 1/rt(1-u²)
Therefore dy/dx = dy/du * du/dx
= 1/rt(1-u²) * sec²
= sec²x / rt(1-tan²x)
Now, at x = 0, sec²0 = 1
and rt(1-tan²0) = rt(1-0) = rt(1) = 1
therefore dy/dx = 1/1 = 1

 

[Riviet]

still wrong: the correct answer is root two... 1-tan^2x does not equal sec^2x. it equals 2-sec^2.

someone please agree with me otherwise youse all will shake my foundation of mathematical learning... lol!

Yes, you are right that 1-tan²x=2-sec²x, but that is not the point of the question. Even if we use 2-sec²x instead:

When x=0
dy/dx=sec²0/rt(2-sec²0)
=1/rt(2-1), since sec²0=1/cos²0=1/1=1
=1

 

[Porcia]

thanks riviet! before SoulSearcher edited his post he had somehing like
d/dx{arcsin(tan x)} = sec2x[1/rt(1-tan2x)]
= sec2x/rt(sec2x)
which didnt make sense... but now i get it... its funny- out of everyone here only Riviet's method is correct.,..hahaha

 

[SoulSearcher]

I didn't really edit it, just said that I screwed it up in that particular post

 

[pLuvia]

thanks riviet! before SoulSearcher edited his post he had somehing like
d/dx{arcsin(tan x)} = sec2x[1/rt(1-tan2x)]
= sec2x/rt(sec2x)
which didnt make sense... but now i get it... its funny- out of everyone here only Riviet's method is correct.,..hahaha

Just missed the square root sign