Question #1506a

2 Answers
Nov 21, 2017

dy/dx = 1/sqrt(1+x^2)

Explanation:

You can simply use the chain rule:

d/dx ln(x+sqrt(1+x^2)) = (1/(x+sqrt(1+x^2)) )d/dx(x+sqrt(1+x^2))

d/dx ln(x+sqrt(1+x^2)) = (1/(x+sqrt(1+x^2)) ) (1+x/sqrt(1+x^2))

d/dx ln(x+sqrt(1+x^2)) = (1/(x+sqrt(1+x^2)) ) (x+sqrt(1+x^2))/sqrt(1+x^2)

d/dx ln(x+sqrt(1+x^2)) = 1/sqrt(1+x^2)

or you can note that:

f(x) = ln(x+sqrt(1+x^2))

is the logarithmic form of the inverse hyperbolic sine function so that:

y= ln(x+sqrt(1+x^2)) <=> sinhy = x

differentiating the second expression implicitly:

d/dx (sinhy) = 1

coshy dy/dx = 1

dy/dx = 1/coshy

and from the identity:

cosh^2y-sinh^2y = 1

we then get:

coshy = sqrt(1+sinh^2y) = sqrt(1+x^2)

and:

dy/dx = 1/sqrt(1+x^2)

Nov 21, 2017

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