Q.322. Differentiation.

Question 322.
Find dy/dx if y = arctan [x + √(1 + x^2)].

Answer 322.
Let x = tanu

=> x + √(1 + x^2)
= tanu + √(1 + tan^2 u)
= tanu + secu
= (sinu + 1) / cosu
= [sin(u/2) + cos(u/2)]^2 / [cos^2 (u/2) - sin^2 (u/2)]
= [sin(u/2) + cos(u/2)] / (cos(u/2) - sin(u/2)]
= [1 + tan(u/2)] / [1 - tan(u/2)]
= tan(π/4 + u/2)

=> y
= arctan [tan(π/4 + u/2)]
= (π/4 + u/2)
= π/4 + (1/2) arctanx
 => dy/dx
= 1 / [2(1 + x^2)].

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