How to take the derivative of $Y=\log(x+\sqrt{a^2+x^2})$?

Question

$$Y=\log(x+\sqrt{a^2+x^2})$$ Find $\dfrac {dY}{dx}$.

My answer:

$$\dfrac 1{x+\sqrt{a^2+x^2}}\cdot\frac{d}{dx}\sqrt{a^2+x^2})$$ Which again goes to a chain rule as $\frac{2}{3} \sqrt{a^2+x^2}\cdot \frac{d}{dx} (a^2+x^2)$

Is there an alternative way other than this? Hints, please.

Answer 1

I think you missed some terms $$Y=\log(x+\sqrt{a^2+x^2})$$ Let $f=x+\sqrt{a^2+x^2}$ which makes $Y=\log(f)$. So $$\frac{dY}{dx}=\frac{dY}{df} \times \frac{df}{dx}=\frac 1 f\times \left(1+\frac{x}{\sqrt{a^2+x^2}}\right)=\frac 1 f\times \left(\frac{x+\sqrt{a^2+x^2}}{\sqrt{a^2+x^2}}\right)$$ Replacing $f$ by its expression finally leads to $$\frac{dY}{dx}=\frac{1}{\sqrt{a^2+x^2}}$$

Answer 2

Let $f(x)$ and $g(x)$ be differentiable .When $g(x)>0$ let $h(x)=f(x)+(g(x))^{1/2}$.When $h(x)>0$ we have $$\frac {d}{dx} \log h(x)=\frac {1}{h(x)}h'(x).$$ And we have $$h'(x)=f'(x)+(1/2)(g(x))^{-1/2}g'(x).$$ It may be possible to simplify the resulting mess, depending on what $f$ and $g$ are. And no, there isn't any other way to do this unless $h$ can be simplified before differentiating.