`y = arctan(sqrt((1 - x)/(1 + x)))` Find the derivative of the function. Simplify where possible.

Textbook Question

Chapter 3, 3.5 - Problem 60 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`d/(dt) tan^-1(t)=1/(1+t^2)`

`y=tan^-1(sqrt((1-x)/(1+x)))`

`y'=(1)/(1+((1-x)/(1+x))) * d/(dx)sqrt((1-x)/(1+x))`

`y'=((1+x)/(1+x+1-x)) * d/(dx) (1-x)^(1/2) (1+x)^(-1/2)`

`y'=((1+x)/2) * ((1-x)^(1/2) d/(dx)(1+x)^(-1/2) +(1+x)^(-1/2) d/(dx) (1-x)^(1/2))`

`y'=((1+x)/2)*((1-x)^(1/2) (-1/2)(1+x)^(-3/2) + (1+x)^(-1/2)(1/2)(1-x)^(-1/2)(-1))`

`y'=((1+x)/2) * ((-(1-x)^(1/2))/(2(1+x)^(3/2)) + (-1)/(2(1+x)^(1/2)(1-x)^(1/2)))` 

`y'=((1+x)/2) * ((-(1-x)-(1+x))/(2(1+x)(1+x)^(1/2)(1-x)^(1/2)))`

`y'=((1+x)/2) *((-1)/((1+x)(1+x)^(1/2)(1-x)^(1/2)))`

`y'=(-1)/(2(1+x)^(1/2)(1-x)^(1/2))`

`y'=(-1)/(2sqrt(1-x^2))`

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