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Question #e6d02

1 Answer

Nov 5, 2016

$\frac{d y}{d x} = \frac{1 - ln (x)}{x^{2}}$ $dy/dx=(1-ln(x))/x^2$

Explanation:

Using the quotient rule

$d/dxf(x)/g(x) = (g(x)f'(x)-f(x)g'(x))/[g(x)]^2$

along with the derivatives

$d/dxln(x) = 1/x$
$d/dxx = 1$

we have

$dy/dx = d/dxln(x)/x$

$=(x(d/dxln(x))-ln(x)(d/dxx))/x^2$

$=(x(1/x)-ln(x)(1))/x^2$

$=(1-ln(x))/x^2$

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