Differentiate the following functions with respect to x?

(1) #sin{2tan^-1[sqrt((1-x)/(1+x))]}#

(2)#cot^-1(sqrt(1+x^2)+x)#

(3)#tan^-1((a+x)/(1-ax))#

1 Answer
Feb 28, 2018

(1) #dy/dx=(-2xy)/(x^2-y^2)#

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

(3) #dy/dx=1/(1+x^2)#

Explanation:

#(1)" " sin{2tan^-1[sqrt((1-x)/(1+x))]}#

Let
#y=sin{2tan^-1[sqrt((1-x)/(1+x))]}#

Simplifying

#x=cos2t, " "sqrt((1-x)/(1+x))=sqrt((1-cos2t)/(1+cos2t))#

#1-cos2t=2sin^2t, " ", 1+cos2t=2cos^2t#

#sqrt((1-cos2t)/(1+cos2t))=sqrt((2sin^2t)/(2cos^2t))=tant#

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

#tan^-1[sqrt((1-x)/(1+x))]=t#

Thus,
#2tan^-1[sqrt((1-x)/(1+x))]=2t#

Now,
#y=sin2t#

The parametric form is

#x=cos2t#
#y=sin2t#

#dy/dx=((dy)/dt)/((dx)/dt)#

#dx/dt=-2sin2t#

#dy/dt=2cos2t#

#dy/dx=(2cos2t)/(-2sin2t)#

#dy/dx=-tan2t#

#tan2t=(2tant)/(1-tan^2t)#

#tant=y/x#

#tan2t=(2y/x)/(1-(y/x)^2)#

#=((2y)/x)/((x^2-y^2)/(x^2))#

#tan2t=(2xy)/(x^2-y^2)#

#dy/dx=-tan2t#

#dy/dx=(2xy)/(x^2-y^2)#

#(2)" "cot^-1(sqrt(1+x^2)+x)#

Let

#y=cot^-1(sqrt(1+x^2)+x)#

#coty=sqrt(1+x^2)+x#

#(coty)-x=sqrt(1+x^2)#

#(coty-x)^2=1+x^2#
Differentiating wrt x

#2(coty-x)(-csc^2ydy/dx-1)=2x#

#(x-coty)(1+csc^2ydy/dx)=x#

#1+csc^2ydy/dx=x/(x-coty)#

#csc^2ydy/dx=x/(x-coty)-1#

#csc^2ydy/dx=(x-(x-coty))/(x-coty)#
#csc^2ydy/dx=(coty)/(x-coty)#

#dy/dx=1/csc^2y((coty)/(x-coty))#

#csc^2y=1+cot^2y#

#dy/dx=1/(1+cot^2y)((-coty)/(coty-x))#

#coty=sqrt(1+x^2)+x#

#coty-x=sqrt(1+x^2)#

#cot^2y=(sqrt(1+x^2)+x)^2#

#cot^2y=1+x^2+2sqrt(1+x^2)+x^2#

#1+cot^2y=1+1+x^2+x^2+2sqrt(1+x^2)#

#=2+2x^2+2sqrt(1+x^2)#

#=2(1+x^2)+2sqrt(1+x^2)#
#1+cot^2y=2(1+x^2+sqrt(1+x^2))#

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

#(3)" " tan^-1((a+x)/(1-ax))#
Let
#y=tan^-1((a+x)/(1-ax))#

#tany=(a+x)/(1-ax)#

#(1-ax)tany=a+x#
Differentiating wrt x

#(1-ax)sec^2ydy/dx+tany(0-a(1))=0+1#

#(1-ax)sec^2ydy/dx-atany=1#

#(1-ax)sec^2ydy/dx=1+atany#

#sec^2ydy/dx=(1+atany)/((1-ax))#

#dy/dx=((1+atany)/((1-ax)))/sec^2y#

#sec^2y=1+tan^2y#

#dy/dx=((1+atany)/((1-ax)))/(1+tan^2y)#

#dy/dx=(1+atany)/((1+tan^2y)(1-ax))#

#tany=(a+x)/(1-ax)#
#tan^2y=((a+x)/(1-ax))^2#

#dy/dx=(1+a((a+x)/(1-ax)))/((1+(((a+x)/(1-ax))^2))(1-ax))#

#dy/dx=(((1-ax+a(a+x)))/(1-ax))/((1-ax)((1-ax)^2+(a+x)^2)/(1-ax)^2)#

#dy/dx=(1-ax+a^2+ax)/((1-ax)^2+(a+x)^2)#

#dy/dx=(1+a^2)/(1-2ax+a^2x^2+a^2+2ax+x^2)#

#dy/dx=(1+a^2)/(1+a^2x^2+a^2+x^2)#

#dy/dx=(1+a^2)/(1+a^2(x^2+1)+x^2)#

#dy/dx=(1+a^2)/(1+x^2+a^2(x^2+1))#

#dy/dx=(1+a^2)/((1+x^2)(1+a^2))#

#dy/dx=1/(1+x^2)#