# Question 4d1ca

Jun 30, 2017

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

#### Explanation:

Two things are required to solve this problem. The chain rule, and knowledge of the derivative of $\ln \left(x\right)$. The chain rule states that,

"d"/("d"x) "f"("g"(x)) = "g"'(x)"f"'("g"(x)).

And the derivative of $\ln \left(x\right)$ is $\frac{1}{x}$.

Then,

"d"/("d"x) ln(x+sqrt(x^2+1)) = ("d"/("d"x) (x+sqrt(x^2+1))) * 1/(x+sqrt(x^2+1)).

Then the derivative of $x + \sqrt{{x}^{2} + 1}$ needs to be computed.

"d"/("d"x) (x+(x^2+1)^(1/2)) = 1 + 1/2 (x^2+1)^(-1/2) *2x,
"d"/("d"x) (x+(x^2+1)^(1/2)) = 1 + x/sqrt(x^2+1),
"d"/("d"x) (x+(x^2+1)^(1/2)) = (sqrt(x^2+1) + x)/(sqrt(x^2+1)).

Then, substituting,

"d"/("d"x) ln(x+sqrt(x^2+1)) = (sqrt(x^2+1) + x)/(sqrt(x^2+1)) * 1/(x+sqrt(x^2+1)).

The factor of $x + \sqrt{{x}^{2} + 1}$ cancels.

"d"/("d"x) ln(x+sqrt(x^2+1)) = 1/sqrt(x^2+1)#.

Incidentally, the function $\ln \left(x + \sqrt{{x}^{2} + 1}\right)$ is the inverse hyperbolic sine function. You can read more about hyperbolic functions and their derivatives here .