Derivative of natural log with chain rule. Is there a better way?

Question

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?

Answer 1

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 $$f'(x)={\frac {1}{\sqrt {{x}^{2}-1}}}$$

Answer 2

Keep going! $$\begin{align}\frac{1}{x + \sqrt{x^2 - 1}}\left(1 + \frac{2x}{2\sqrt{x^2 - 1}}\right)&=\frac{1}{x + \sqrt{x^2 - 1}}\left(\frac{x+\sqrt{x^2-1}}{\sqrt{x^2 - 1}}\right)\\&=\frac1{\sqrt{x^2-1}}\end{align}$$

Answer 3

\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*}

Answer 4

$$\frac{1}{x + \sqrt{x^2 - 1}} \cdot (1 + \frac{2x}{2 \cdot \sqrt{x^2 - 1}})$$ $$=\frac{1}{x + \sqrt{x^2 - 1}}\frac{x + \sqrt{x^2 - 1}}{\sqrt{x^2 - 1}}$$ $$=\frac{1}{\sqrt{x^2 - 1}}$$

Answer 5

Here is a sneaky way to deal with one of those derivatives. Notice that $h(x)=\cosh^{-1}(x)$, therefore $$ h’(x)=\frac{1}{\sinh\left(\cosh^{-1}(x)\right)} = \frac{1}{\sqrt{\cosh^2\left(\cosh^{-1}(x)\right)-1}}=\frac{1}{\sqrt{x^2-1}}. $$

Answer 6

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: $$h'(x)=\frac{1}{\sqrt{x^2-1}}$$