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

2 Answers
Jim G.
May 4, 2018

#1/(sqrt((1+x)(1-x)^3))#

Explanation:

#"differentiate using "color(blue) "quotient/chain rules"#

#"given "y=(g(x))/(h(x))" then"#

#dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

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

#g(x)=(1+x)^(1/2)rArrg'(x)=1/2(1+x)^(-1/2)#

#h(x)=(1-x)^(1/2)rArrh'(x)=-1/2(1-x)^(-1/2)#

#rArrdy/dx=(1/2(1-x)^(1/2)(1+x)^(-1/2)+1/2(1+x)^(1/2)(1-x)^(-1/2))/(1-x)#

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

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

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

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

P dilip_k
May 4, 2018

#y=sqrt(1+x)/sqrt(1-x)#

Let #x=cos2theta#
#(dx)/(d theta)=-2sin2theta#

So

#y=sqrt(1+cos2theta)/sqrt(1-cos2theta)#

#=>y=sqrt(2cos^2theta)/sqrt(2sin^2theta)=cottheta#

#(dy)/(d x)=-csc^2theta*(d theta)/(dx)#

#=-1/sin^2theta*1/(-2sin2theta)#

#=1/(2sin^2theta)*1/(sin2theta)#

#=1/(1-cos2theta)*1/(sqrt(1-cos^2 2theta)#

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

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

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