How do you find the derivative of $f(x)=ln(2x^2+1)$ ?

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Answer 1 · 2016-11-07T11:58:50

$(4x)/(2x^2+1)$

"Function Composition"

$(f(g(x)))'= f’(g(x))g’(x)$

$g(x)' = (2x^2+1)'= 4x$

$f'(ln() ) = 1/g(x)= 1/(2x^2+1)$

$(f(g(x)))'= f’(g(x))g’(x) = 1/(2x^2+1) * 4x= (4x)/(2x^2+1)$

How do you find the derivative of f(x)=ln(2x^2+1)f(x)=ln(2x^2+1)?