Derivative: arctan(sqrt((1-x)/(1+x)))

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In summary, to find the derivative of arctan√((1-x)/(1+x)), we use the chain rule and quotient rule. After simplifying, the derivative is equal to 1/(1 + (sqrt(quotient in radical))^2) * 1/2(quotient in radical)^(-1/2) * (derivative of quotient inside radical).
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Homework Statement


Find the derivative:
arctan[tex]\sqrt{(1-x)/(1+x)}[/tex]


Homework Equations


Chain Rule, Quotient Rule


The Attempt at a Solution


I've gotten to a point where I feel like I either don't know how to simplify from this point, or I've done something wrong:
[1/(1+((1-x)/(1+x)))] * [(-[tex]\sqrt{(1+x)}[/tex]/2*[tex]\sqrt{1-x}[/tex])-([tex]\sqrt{1-x}[/tex]/2*[tex]\sqrt{1+x}[/tex])]
 
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  • #2
It looks about right (I didn't check that closely), but I don't understand some of what you have shown, namely the parts where it looks like you're dividing by 2.

Your derivative should look like this:
1/(1 + (sqrt(quotient in radical))^2) * 1/2(quotient in radical)^(-1/2) * (derivative of quotient inside radical)

The first factor you show looks fine, but I don't understand what you have after that.
 

What is the derivative of arctan(sqrt((1-x)/(1+x)))?

The derivative of arctan(sqrt((1-x)/(1+x))) is equal to -12(1-x)-1(1+x)-1(1+(1-x)2).

How do you simplify arctan(sqrt((1-x)/(1+x)))?

To simplify arctan(sqrt((1-x)/(1+x))), you can use the identity arctan(sqrt(x)) = ½arcsin(x). Therefore, arctan(sqrt((1-x)/(1+x))) can be simplified to ½arcsin((1-x)/(1+x)).

What is the domain of arctan(sqrt((1-x)/(1+x)))?

The domain of arctan(sqrt((1-x)/(1+x))) is all real numbers except -1 and 1, since these values would result in a division by zero in the denominator.

How do you find the critical points of arctan(sqrt((1-x)/(1+x)))?

To find the critical points of arctan(sqrt((1-x)/(1+x))), you can take the derivative and set it equal to 0. This will give you the critical points at x = ½ and x = -½.

What is the graph of arctan(sqrt((1-x)/(1+x)))?

The graph of arctan(sqrt((1-x)/(1+x))) is a continuous, increasing function with a horizontal asymptote at y = ½ and vertical asymptotes at x = -1 and x = 1. It also has a local minimum at x = -½ and a local maximum at x = ½.

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