Question #1506a

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Nov 21, 2017

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Answer 1 · 2017-11-21T23:33:51

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

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)#