Find the derivative and simplify $y = \ln(x+\sqrt{x^2 + a^2})$ I have worked through this problem, but am not sure of my answer.
I am supposed to find the derivative of $y=\ln(x+\sqrt{x^2+a^2})$, $a$ being a constant.
I used the property of natural log to find the derivative of $1$ over $x$ times derivative $x$ and then the chain rule on the square root function.
After simplifying the answer I ended up with is $dy/dx=1/(x^2 + a^2)$... does this seem logical?
 A: The calculation was probably right almost to the end. We use the Chain Rule. Our function has the shape $\ln(g(x))$, where $g(x)=x+\sqrt{x^2+a^2}$.   The derivative is therefore $\frac{g'(x)}{g(x)}$.
We have
$$g'(x)=1+\frac{x}{\sqrt{x^2+a^2}}=\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}.$$
When we divide by $x+\sqrt{x^2+a^2}$ and simplify, we get $\frac{1}{\sqrt{x^2+a^2}}$.
A: I solved this question in two different ways. Andre's solution is more natural and best.

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*(By implicit differentiation)

$$e^y=x+\ln\sqrt{x^2+a^2}\implies e^y y'=1+\frac{x}{\sqrt{x^2+a^2}}\implies e^y y'=\frac{e^y}{\sqrt{x^2+a^2}}\implies y'=\frac{1}{\sqrt{x^2+a^2}}.$$


*(By parametric differention)

Let $x=a\tan\theta$ then $\frac{dx}{d\theta}=a\sec^2\theta$ and also $y=\ln a+\ln|\tan\theta+sec\theta|$. Did you remembered the integral of the secant function? Then $\frac{dy}{d\theta}=\sec\theta$. Hence,
$$y'=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{\cos\theta}{a}=\frac{\frac{a}{\sqrt{x^2+a^2}}}{a}=\frac{1}{\sqrt{x^2+a^2}}.$$
