Derivative of natural log with chain rule. Is there a better way? I'm a bit stuck on taking these two derivatives:
$$h(x) = \ln(x + \sqrt{x^2-1})$$
\begin{align}
h'(x) &= \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{d }{dx} (x + \sqrt{x^2 -1} )\\
&= \frac{1}{x + \sqrt{x^2 - 1}} \biggl(1 + \frac{2x}{2\sqrt{x^2 - 1}}\biggr)
\end{align}
I'm a bit stuck on how to simplify from here? Was there a simpler way somewhere?
Also this one is giving me problems. I am thinking of changing the log forms first?:
$$\ln \frac{(2y+1)^5}{\sqrt{y^2 + 1}}= \ln(2y+1)^5 - \ln \sqrt{y^2 +1}=5 \ln(2y+1) - \frac{1}{2}\,\ln (y^2 +1)$$
so:
$$G'y = \frac{5}{2y+1} \cdot 2 - \frac{2y}{2(y^2 + 1)} = \frac{10}{2y+1} - \frac{y}{2 (y^2 +1)}$$
Is that right?
 A: Keep going! $$\begin{align}\frac{1}{x + \sqrt{x^2 - 1}}\left(1 + \frac{2x}{2\sqrt{x^2 - 1}}\right)&=\frac{1}{x + \sqrt{x^2 - 1}}\left(\frac{x+\sqrt{x^2-1}}{\sqrt{x^2 - 1}}\right)\\&=\frac1{\sqrt{x^2-1}}\end{align}$$
A: by the chain rule we get $$f'(x)={\frac {1}{x+\sqrt {{x}^{2}-1}} \left( 1+{\frac {x}{\sqrt {{x}^{2}-1}}
} \right) }
$$
simplifying we obtain $$f'(x)={\frac {1}{\sqrt {{x}^{2}-1}}}$$
A: \begin{align*}
f(y)&=\ln\dfrac{(2y+1)^{5}}{\sqrt{y^{2}+1}}\\
&=5\ln(2y+1)-\dfrac{1}{2}\ln(y^{2}+1),
\end{align*}
so
\begin{align*}
f'(y)=\dfrac{10}{2y+1}-\dfrac{y}{y^{2}+1}.
\end{align*}
A: $$\frac{1}{x + \sqrt{x^2 - 1}} \cdot (1 + \frac{2x}{2 \cdot \sqrt{x^2 - 1}})$$
$$=\frac{1}{x + \sqrt{x^2 - 1}}\frac{x + \sqrt{x^2 - 1}}{\sqrt{x^2 - 1}}$$
$$=\frac{1}{\sqrt{x^2 - 1}}$$
A: Here is a sneaky way to deal with one of those derivatives. Notice that $h(x)=\cosh^{-1}(x)$, therefore
$$
h’(x)=\frac{1}{\sinh\left(\cosh^{-1}(x)\right)}
= \frac{1}{\sqrt{\cosh^2\left(\cosh^{-1}(x)\right)-1}}=\frac{1}{\sqrt{x^2-1}}.
$$
A: You can also do it this way:
$$h(x)=\log(x+\sqrt{x^2-1})$$
$$\exp(h(x))=x+\sqrt{x^2-1}$$
Now we can differentiate both sides with respect to $x$:
$$\exp(h(x))*h'(x)=1+\frac{x}{\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}=\frac{\exp(h(x))}{\sqrt{x^2-1}}$$
So:
$$h'(x)=\frac{1}{\sqrt{x^2-1}}$$
