If #y = tan^-1((ax - b)/(bx + a))#. Then how will you prove that #dy/dx = 1/(1 + x^2)#??

1 Answer
Steve M
Jul 24, 2017

We have:

# y = tan^(-1)( (ax-b)/(bx+a) ) #

Using # d/dx tan^(-1)x = 1/(1+x^2) #, along with the chain rule we have:

# dy/dx = 1/(1+((ax-b)/(bx+a))^2) * d/dx((ax-b)/(bx+a))#

# " " = 1/(1+((ax-b)/(bx+a))^2) * ( (bx+a)(a) - (ax-b)(b) ) / (bx+a)^2 #

# " " = 1/(((bx+a)^2+(ax-b)^2)/(bx+a)^2) * ( a(bx+a) - b(ax-b) ) / (bx+a)^2#

# " " = (bx+a)^2/((bx+a)^2+(ax-b)^2) * ( a(bx+a) - b(ax-b) ) / (bx+a)^2#

# " " = 1/((bx+a)^2+(ax-b)^2) * ( a(bx+a) - b(ax-b) ) #

# " " = ( a(bx+a) - b(ax-b) )/((bx+a)^2+(ax-b)^2) #

# " " = ( abx+a^2 - abx+b^2 )/(b^2x^2+2abx+a^2+a^2x^2-2abx+b^2) #

# " " = ( a^2 +b^2 )/(a^2+a^2x^2+b^2+b^2x^2) #

# " " = ( a^2 +b^2 )/(a^2(1+x^2)+b^2(1+x^2)) #

# " " = ( a^2 +b^2 )/((a^2+b^2)(1+x^2)) #

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

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