Question #042b9

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Answer 1 · 2017-02-04T20:49:05

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

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

Take the natural logarithm of both sides:

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

The right-hand side can be rewritten using $ln(a^b)=bln(a)$ :

$ln(f(x))=xln(2x)$

Differentiate both sides of the equation. The chain and product rules will be used.

$1/f(x)f'(x)=ln(2x)(d/dxx)+x(d/dxln(2x))$

$1/(2x)^xf'(x)=ln(2x)+x(1/(2x))(d/dx2x)$

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

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