\eqalign{
& y = \sqrt x \cdot \cos (x^2 ) \cdot \ln (x) \cr
& y = x^{\frac{1}
{2}} \cdot \cos (x^2 ) \cdot \ln (x) \cr
& \ln (y) = \ln \left( {x^{\frac{1}
{2}} \cdot \cos (x^2 ) \cdot \ln (x)} \right) \cr
& \ln (y) = \ln \left( {x^{\frac{1}
{2}} } \right) + \ln \left( {\cos (x^2 )} \right) + \ln \left( {\ln (x)} \right) \cr
& \ln (y) = \frac{1}
{2}\ln \left( x \right) + \ln \left( {\cos (x^2 )} \right) + \ln \left( {\ln (x)} \right) \cr
& \frac{1}
{y} \cdot y' = \frac{1}
{{2x}} - 2x \cdot \tan (x^2 ) + \frac{1}
{{x \cdot \ln (x)}} \cr
& y' = \left( {\frac{1}
{{2x}} - 2x \cdot \tan (x^2 ) + \frac{1}
{{x \cdot \ln (x)}}} \right) \cdot \sqrt x \cdot \cos (x^2 ) \cdot \ln (x) \cr
& y' = \frac{1}
{{2x}} \cdot \sqrt x \cdot \cos (x^2 ) \cdot \ln (x) - 2x \cdot \tan (x^2 ) \cdot \sqrt x \cdot \cos (x^2 ) \cdot \ln (x) + \frac{1}
{{x \cdot \ln (x)}} \cdot \sqrt x \cdot \cos (x^2 ) \cdot \ln (x) \cr
& y' = \frac{{\cos (x^2 ) \cdot \ln (x)}}
{{2\sqrt x }} - 2x\sqrt x \cdot \sin (x^2 ) \cdot \ln (x) + \frac{{\cos (x^2 )}}
{{\sqrt x }} \cr
& y' = \frac{{\cos (x^2 ) \cdot \ln (x) - 4x^2 \sin (x^2 ) \cdot \ln (x) + 2\cos (x^2 )}}
{{2\sqrt x }} \cr}
$
Help dat?
Zie Logaritmisch differentiëren [http://www.hhofstede.nl/modules/logdiff.htm]
WvR
16-5-2015