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