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]

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

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

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

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

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

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

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

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