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

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

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