ln and Square Root Diff

[Jason76]

\(\displaystyle f(x) = \ln[x + \sqrt{x^{2} - 9}]\)

\(\displaystyle f(x) = \ln[x + (x^{2} - 9)^{1/2}]\)

\(\displaystyle f'(x) = \ln[1 + \dfrac{1}{2} (u)^{-1/2} (2)]\)

\(\displaystyle f'(x) = \ln[1 + \dfrac{1}{2} (x^{2} - 9)^{-1/2} (2)]\)

 

[lookagain]

\(\displaystyle f(x) = \ln[x + \sqrt{x^{2} - 9}]\)

\(\displaystyle f(x) = \ln[x + (x^{2} - 9)^{1/2}]\)

\(\displaystyle f'(x) = \ln[1 + \dfrac{1}{2} (u)^{-1/2} (2)]\)

\(\displaystyle f'(x) = \ln[1 + \dfrac{1}{2} (x^{2} - 9)^{-1/2} (2)]\)

No, it's 1 divided by the quantity you're taking the logarithm of, multiplied by the derivative of that quantity.


\(\displaystyle d[ln(u)] \ = \ \dfrac{du}{u}\)

 

[Jason76]

\(\displaystyle f(x) = \ln[x + \sqrt{x^{2} - 9}]\)

\(\displaystyle f(x) = \ln[x + (x^{2} - 9)^{1/2}]\)

\(\displaystyle f'(x) = \dfrac{1}{u}[\dfrac{d}{dx} u]\)

\(\displaystyle f'(x) = \dfrac{1}{u}[1 + ((v)(\dfrac{d}{dx} v))]\)

\(\displaystyle f'(x) = \dfrac{1}{u}[1 + (\dfrac{1}{2}(v)^{-1/2}(2))]\)

\(\displaystyle f'(x) = \dfrac{1}{u}[1 + (\dfrac{1}{2}(x^{2} - 9)^{-1/2}(2))]\)

\(\displaystyle f'(x) = \dfrac{1}{u}[1 + (x^{2} - 9)^{-1/2}]\)

\(\displaystyle f'(x) = \dfrac{1}{x + (x^{2} - 9)^{1/2}}[1 + (x^{2} - 9)^{-1/2}]\)