What are the first and second derivatives of $f (x) = e^{2 x} (ln x)$ $f(x)=e^(2x)(lnx)$ ?

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Answer 1 · 2018-08-08T13:04:48

$f^'(x)= e^(2x)( 1/x+ 2(lnx))$
$f^''(x)= (e^(2x)(2x-1))/ x^2$
$+ 2 e^(2x) ( 1/x+ 2 *(lnx))$

$f(x)= e^(2x)(ln x)$

$f^'(x)= e^(2x)* 1/x+ 2e^(2x)*(lnx)$

$f^'(x)= e^(2x)( 1/x+ 2(lnx))$

$f^''(x)= (x *2e^(2x)-e^(2x)*1)/ x^2$

$+ 2(e^(2x) * 1/x+ 2e^(2x) *(lnx))$

$f^''(x)= (e^(2x)(2x-1))/ x^2$

$+ 2 e^(2x) ( 1/x+ 2 *(lnx))$ [Ans]

What are the first and second derivatives of f(x)=e^(2x)(lnx) f(x)=e2x(lnx)f(x)=e^(2x)(lnx) ?