Question

If y = `log(x + sqrt(x^2 + 4))`, show that `dy/dx = 1/sqrt(x^2 + 4)`

Answer 1

y = `log(x + sqrt(x^2 + 4))`

Differentiate w.r.t.x

`dy/dx = 1/(x + sqrt(x^2 + 4)).d/dx(x + sqrt(x^2 + 4))`

= `1/((x + sqrt(x^2 + 4)))[1 + (2x)/(2sqrt(x^2 + 4))]`

= `1/((x + sqrt(x^2 + 4)))[(sqrt(x^2 + 4) + x)/sqrt(x^2 + 4)]`

= `1/sqrt(x^2 + 4)`