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

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Answer 1 · 2015-06-05T20:27:30

I got:

#-(1)/(2sqrt(1-x^2))#

The derivative of #arctanu# is #(1/(1+(u(x))^2))((du(x))/(dx))#.

So since #u(x) = sqrt((1-x)/(1+x))#:

#d/(dx)(arctansqrt((1-x)/(1+x)))#

# = 1/(1+(1-x)/(1+x))*(1/(2sqrt((1-x)/(1+x))))*[((1+x)*(-1) - (1-x)*(1))/(1+x)^2]#

You can see there are multiple chain rules here.

# = [(-1cancel(-x)-1cancel(+x))/(1+x)^2][1/(2sqrt((1-x)/(1+x))(1+(1-x)/(1+x)))]#

Multiply in the #sqrt((1-x)/(1+x))# and cancel out the #2#:

# = [(-cancel(2))/(1+x)^2][1/(cancel(2)(sqrt((1-x)/(1+x))+((1-x)/(1+x))^"3/2"))]#

Merge the fractions:

# = -(1)/((1+x)^"4/2"(sqrt((1-x)/(1+x))+((1-x)/(1+x))^"3/2"))#

Multiply in the #(1+x)^"4/2"#:

# = -(1)/(sqrt(1-x)*(1+x)^"3/2"+(1-x)^"3/2"*sqrt(1+x)#

Simplify by turning #sqrt(1-x)# and #sqrt(1+x)# into #sqrt(1-x^2)#:

# = -(1)/((1-x)^"1/2"(1+x)^"1/2"(1+x)+(1-x)(1-x)^"1/2"(1+x)^"1/2"#

Factor out the #sqrt(1-x^2)# and cancel what's now inside:

# = -(1)/(sqrt(1-x^2)*(1+x)+(1-x)*sqrt(1-x^2))#

# = -(1)/(sqrt(1-x^2)(1cancel(+x)+1cancel(-x)))#

# = color(blue)(-(1)/(2sqrt(1-x^2)))#

Wolfram Alpha gives the derivative as (assuming #x > 0#)

# = -(1)/(2sqrt(1-x^2))#,

and in general, gives

#sqrt((1-x)/(1+x))/(2(1-x))#.