Topic: Differentiating y = tan(arcsin x)

[JaWiB]

Where am I going wrong on this one? Both my calculator and the book give me y' = (1-x^2)^(-3/2) but I get y' = (1-x^2)^(-1/2)

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y = tan(arcsin x)
sin y = x   (draw my little triangle here)
tan y = x/sqrt(1-x^2)  (differentiate)
sec^2 y * y' = ( sqrt(1-x^2) - (-x^2)(1-x^2)^(-1/2) ) / (1-x^2)
sec^2 y * y' = (1-x^2)^(-1/2) ( 1 - x^2 + x^2 ) /  (1-x^2)
sec^2 y * y' = 1/(1-x^2)^(3/2)  (substitute sec^2 y = 1/sqrt(1-x^2) )
1/(1-x^2) * y' = 1/(1-x^2)^(3/2)
y' = 1/sqrt(1-x^2)

Here's the triangle I used:

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1 /| 
 / | x
/y |
sqrt(1-x^2)


[Sang-drax]

I cannot follow your calculations, but here's my solution:

The derivative of f(x) = tan(x) is
f'(x) = 1 - tan^2(x) = 1 - sin^2(x) / (1 - sin^2(x))

The derivative of g(x) = arcsin(x) is
g'(x) = 1 / sqrt(1 - x^2)


Since y = f(g(x)) the derivative of y is
y' = f'(g(x)) * g'(x) =
(1 - x^2/(1-x^2) * 1 / sqrt(1-x^2)

Simplify and you get the answer from the book.

[Sang-drax]

About your solution:
y = tan(arcsin x)  *does not mean that*
sin y = x

[JaWiB]

The derivative of f(x) = tan(x) is
f'(x) = 1 - tan^2(x) = 1 - sin^2(x) / (1 - sin^2(x))

Are you sure that's right?
d/dx[tan(x)] = sec^2(x) doesn't it?

And sec^2(x) = 1 + tan^2(x) I thought?

But I think I see what I did wrong. Something the teacher said confused me, anyways I'll give it another try later.

[Sang-drax]

Are you sure that's right?

Absolutely.
tan^2(x) = sin^2(x) / cos^2(x)  = sin^2(x) / (1 - sin^2(x))

BTW, I am not familiar with the sec function.

[Mike]

sec(x) = 1 / cos(x) http://mathworld.wolfram.com/Secant.html
csc(x) = 1 / sin(x) http://mathworld.wolfram.com/Cosecant.html

Sang: d/dx [tan^2(x)] = sec^2(x) = 1 + tan^2(x)  whereas you have a minus.

[Sang-drax]

Oh, sorry.