How do you differentiate #f(x) = sin ( x² ln(x) )#?

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Answer 1 · 2016-01-29T15:55:51

#f'(x)=(2xln(x)+x)cos(x^2ln(x))#

To find the derivative of a sine function like this, we will have to use the chain rule:

#d/dx(sin(u))=cos(u)*u'#

In this instance, #u=x^2ln(x)#, so differentiation yields

#f'(x)=cos(x^2ln(x))*d/dx(x^2ln(x))#

To differentiate #x^2ln(x)#, the product rule is necessary. Recall that the derivative on #ln(x)# is #1/x#.

#d/dx(x^2ln(x))=ln(x)d/dx(x^2)+x^2d/dx(ln(x))=2xlnx+x#

Plugging this back in, we see that

#f'(x)=(2xln(x)+x)cos(x^2ln(x))#