# Differentiation using product rule

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• Apr 21st 2007, 02:30 AM
dadon
Differentiation using product rule
Hey all,

I have done this question but don't seem to get the same answer as given, i don't know where I am going wrong? can someone check for me please. Thank you

Find dy/dx using the product rule when: y=x^2lnxcos3x

u=x^2
du/dx=2x

v=lnxcos3x
a=lnx, b=cos3x
dv/dx=a.db/dx+b.da/dx

da/dx=1/x
db/dx=-3sin3x
dv/dx=lnx.-3sin3x+cos3x.1/x
dv/dx=-3lnxsin3x+cos3x/x

dy/dx=u.dv/dx+v.du/dx
dy/dx=x^2.-3lnxsin3x+cos3x/x+lnxcos3x.2x
dy/dx=-3x^2lnxsin3x+co3x/x+2xlnxcos3x
• Apr 21st 2007, 03:55 AM
topsquark
Quote:

Originally Posted by dadon
Hey all,

I have done this question but don't seem to get the same answer as given, i don't know where I am going wrong? can someone check for me please. Thank you

Find dy/dx using the product rule when: y=x^2lnxcos3x

u=x^2
du/dx=2x

v=lnxcos3x
a=lnx, b=cos3x
dv/dx=a.db/dx+b.da/dx

da/dx=1/x
db/dx=-3sin3x
dv/dx=lnx.-3sin3x+cos3x.1/x
dv/dx=-3lnxsin3x+cos3x/x

dy/dx=u.dv/dx+v.du/dx
dy/dx=x^2.-3lnxsin3x+cos3x/x+lnxcos3x.2x
dy/dx=-3x^2lnxsin3x+co3x/x+2xlnxcos3x

The error is in the second to the last line: you didn't distribute the x^2.

How about we write it this way: dy/dx = y' It works out to be a bit neater.

Then we have
y = [x^2]*[ln(x)cos(3x)]

y' = [x^2]' * [ln(x)cos(3x)] + [x^2] * [ln(x)cos(3x)]'

Now
[x^2]' = 2x
and
[ln(x)cos(3x)]' = [ln(x)]' * [cos(3x)] + [ln(x)] * [cos(3x)]'
where
[ln(x)]' = 1/x and [cos(3x)] = -sin(3x) * 3 <-- By the chain rule.

So
[ln(x)cos(3x)]' = (1/x)*cos(3x) - 3*ln(x)*sin(3x)

Thus

y' = [x^2]' * [ln(x)cos(3x)] + [x^2] * [ln(x)cos(3x)]'

y' = 2x*ln(x)*cos(3x) + x^2*[(1/x)*cos(3x) - 3*ln(x)*sin(3x)]

y' = 2x*ln(x)*cos(3x) + x*cos(3x) - 3x^2*ln(x)*sin(3x)

-Dan