I'm supposed to calculate the differential of f(x)= sin(x)ln(x)(x-4)^2 using the product rule. I know what the product rule is but I can't split this into two bits that are easy to differentiate. How do I do it?

You say you are familiar with the product rule i.e. f(x)=u(x)v(x) f'(x)= u(x)v'(x)+v(x)u'(x) (Equation 1)

OK so why don't we try applying that here let's try splitting the function in this problem down into two parts. Let: u(x)=(x-4)^2 v(x)=sin(x)ln(x)

My guess is that v(x) is looking a bit difficult but don't worry we'll get to it. We already have u(x) and v(x) but we need to calculate u'(x) and v'(x). Let's find u'(x):

u(x)=(x-4)^2 u(x)= x^2-8x+16 (multiplied out the brackets) u'(x)=2x-8 (differentiated - multiplied through by the power of x and reduce the power of x by one)

Now we just need v'(x), currently we have:

v(x)=sin(x)*ln(x)

If you look at this, it similar to the problem we had to start with so all we need to do is apply the product rule again. Let: t(x)=sin(x) r(x)=ln(x)

v(x)=t(x)r(x) v'(x)= r'(x)t(x)+t'(x)r(x) (Equation 2)

Try and have a go yourself from here but if you need more help or you've completed the problem and want to check your answer, read on:

We have r(x) and t(x) so let's calculate r'(x) and t'(x):

r(x)= ln(x) r'(x)= x^-1=1/x (recall differential of a log.)

t(x) = sin(x) t'(x) = cos(x) (recall differentials of trig. functions)

Now we can put all the values into Equation 2. We get:

v'(x)= sin(x)/x + cos(x)*ln(x)

Now we've found all the bits for Equation 1 so let's plug in all those values:

f'(x) = (x-4)^2*(sin(x)/x + cos(x)*ln(x))+sin(x)ln(x)(2x-8)