Question

If y = (1/3)log [{x + 1}/{√(x² + x + 1)}] + (1/√3) tan^–1 [(2x – 1)/(√3)] then (dy/dx) =______

(a) [1/(1 + x³)]

(b) [(x² + x + 1)/(x – 1)]

(c) [1/(1 – x³)]

(d) none of these

Answer 1

The correct option (a) [1/(1 + x³)]

Explanation:

y = (1/3) log [(x + 1)/√(x² – x + 1)] + (1/√3) tan^–1 [(2x – 1)/√3]

∴ y = (1/3) [log (x + 1) – log (x² – x + 1)^(1/2)] + (1/√3) tan^–1 [(2x – 1)/(√3)]

∴ (dy/dx) = (1/3) [{1/(x + 1)} – (1/2) ∙ {(2x – 1)/(x² – x + 1)}] + (1/√3) ∙ [1/(1 + {(2x – 1)/√3}²)] ∙ (2/√3)

= (1/3) [{2(x² – x + 1) – (x + 1)(2x – 1)}/{2(x + 1) (x² – x + 1)}] + (2/3) [1/{3 + (2x – 1)²}]

= (1/3) [{3 – 3x}/{2(x + 1)(x² – x + 1)}] + [1/{2(x² – x + 1)}]

= [{(1 – x) + (1 + x)}/{2(x + 1)(x² – x + 1)}]

= [1/{(x + 1)(x² – x + 1)}]

= [1/(1 + x³)]