Question

Find Derivative of f(x) = tan-1 (√1+x² + x) with respect to x.

Answer 1

f(x) = y = tan^-1( sqrt(1+x²) +x )

tany = sqrt(1+x²) +x

differentiating w.r. to x

sec²y (dy/dx) = (2x)/(2sqrt(1+x²) ) + 1 =  x/sqrt(1+x²)  + 1 

                    = [ ( x + sqrt(1+x²))/ sqrt(1+x²) ]... Eqn 1

now sec²y = 1 + tan²y  = 1 + [ sqrt(1+x²) +x ]² = 1 + [(1+x²) + x² + 2x sqrt(1+x²)] 

            = 2x² + 2x sqrt(1+x²) + 2 = 2 [x² + x sqrt(1+x²) + 1 ] = 2sqrt(1+x²)[ sqrt(1+x²) + x]

substituting in Eqn 1

2sqrt(1+x²)[ sqrt(1+x²) + x] (dy/dx) =  [ ( x + sqrt(1+x²))/ sqrt(1+x²) ]

(dy/dx) = 1/[ 2(1+x²)]