On -1 < x < 1?
The answer is apparently 1/2cos(arcsin x/2)...?
The answer is apparently 1/2cos(arcsin x/2)...?
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If y = arcsin(x/2) then x = 2siny and we have:
Take the derivative of both sides with respect to x:
d/dx(x) = d/dx(2siny)
=> 1 = 2dy/dx*d/dy(siny)
=> 1 = 2cosy*dy/dx
=> dy/dx =1/(2cosy)
=> dy/dx = 1/(2cos(arcsin(x/2))
as stated in the question. Is this what you were looking for?
Just seen the answer above mine; yes it is much better expressed like this.
cosy = (1 - sin^2y)^(1/2) = (1 - (x^2)/4)^(1/2)
and so:
dy/dx = 1/(2(1 - (x^2)/4)^(1/2))
This derivative is easier to compute.
***********Additional Info*************************
In fact this result is quite useful for when you come across integrals of the form:
Int(1/sqrt(1 - x^2) dx)
Now you know that it is simply (removing all the '2' multipliers from the above argument):
Int(1/sqrt(1 - x^2) dx) = arcsinx
Make a note of this result as this integral can be tricky to do using traditional techniques like 'substitution' or 'integration by parts', keeping this note tucked away is an excellent shortcut to this type of problem.
Take the derivative of both sides with respect to x:
d/dx(x) = d/dx(2siny)
=> 1 = 2dy/dx*d/dy(siny)
=> 1 = 2cosy*dy/dx
=> dy/dx =1/(2cosy)
=> dy/dx = 1/(2cos(arcsin(x/2))
as stated in the question. Is this what you were looking for?
Just seen the answer above mine; yes it is much better expressed like this.
cosy = (1 - sin^2y)^(1/2) = (1 - (x^2)/4)^(1/2)
and so:
dy/dx = 1/(2(1 - (x^2)/4)^(1/2))
This derivative is easier to compute.
***********Additional Info*************************
In fact this result is quite useful for when you come across integrals of the form:
Int(1/sqrt(1 - x^2) dx)
Now you know that it is simply (removing all the '2' multipliers from the above argument):
Int(1/sqrt(1 - x^2) dx) = arcsinx
Make a note of this result as this integral can be tricky to do using traditional techniques like 'substitution' or 'integration by parts', keeping this note tucked away is an excellent shortcut to this type of problem.
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y = arcsin (x/2)
sin y = x/2
cos y (dy/dx) = 1/2
dy/dx = (1/2) / cos(arcsin (x/2))
BUT, note that the angle whose sine is x/2
has as its cosine the quantity (1-x^2/4),
so dy/dx = 2 / (4-x^2)
Check my arith to make sure my factors of "2" are OK
sin y = x/2
cos y (dy/dx) = 1/2
dy/dx = (1/2) / cos(arcsin (x/2))
BUT, note that the angle whose sine is x/2
has as its cosine the quantity (1-x^2/4),
so dy/dx = 2 / (4-x^2)
Check my arith to make sure my factors of "2" are OK
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http://www.wolframalpha.com/input/?i=dx+…