composite rule and derivative question

[nil101]

Please can you help with this one?

Use the composite rule(chain rule) to show that the function

\(\displaystyle \ h(x) = \ln (x + \sqrt {x^2 {\rm - }1{\rm }} {\rm ) }(x > 1)
\\)

has derivative

\(\displaystyle \
h'(x) = \frac{1}{{\sqrt {x^2 - 1} }}
\\) ?


Thanks for looking.

 

[soroban]

Hello, nil101!

Use the composite rule to show that the function
. . \(\displaystyle h(x) \:=\: \ln\left(x\,+\,\sqrt{x^2 \,-\,1}\right),\;x\,>\.1\)

has derivative: \(\displaystyle \:h'(x) \:=\:\frac{1}{{\sqrt {x^2\,-\,1}}\)

We have: \(\displaystyle \:h(x) \;=\;\ln\left[x\,+\,(x^2\,-\,1)^{\frac{1}{2}}\right]\)


Using the composite rule (chain rule), we have:

\(\displaystyle \L h'(x)\;=\;\frac{1}{x\,+\,\left(x^2\,-\,1\right)^{\frac{1}{2}}} \,\cdot\,\left[1\,+\,\frac{1}{2}(x^2\,-\,1)^{-\frac{1}{2}}\cdot\,2x\right]\)

. . \(\displaystyle \L=\;\frac{1}{x\,+\,\sqrt{x^2\,-\,1}}\cdot\left[1\,+\,\frac{x}{\sqrt{x^2\,-\,1}}\right]\)

. . \(\displaystyle \L=\;\frac{1}{x\,+\,\sqrt{x^2\,-\,1}}\cdot\left[\frac{\sqrt{x^2\,-\,1}\,+\,x}{\sqrt{x^2\,-\,1}}\right]\)

. . \(\displaystyle \L= \;\frac{1}{\sout{x\,+\,\sqrt{x^2\,-\,1}}}\,\cdot\frac{\sout{x\,+\,\sqrt{x^2\,-\,1}}}{\sqrt{x^2\,-\,1}}\)

. . \(\displaystyle \L=\;\frac{1}{\sqrt{x^2\,-\,1}}\)