Can you double check my working please?

Question

This is the question :
Use logarithmic differentiation to find dy/dx ,
y = (x^3) * (sin(x)^2) * (cos(x)^3)
* Here is my working :
y' = x^2 * sin(x) * (cos(x))^2 * (3*sin(x)(cos(x) - x(5*(sin(x))^2) - 2))

y'=-x^(2) * sin(x)(cos(x))^2 *(x*(5*(sin(x))^2-2)-3(sin(x)*cos(x))

The teachers answer is quite different to mine , I have uploaded it to imageshack : http://img408.imageshack.us/img408/6100/questionfa0.jpg

Can someone double check my working and based on the key fron the teachers answer tell me how many marks I would get for it ?

Second line got cut off:
y'=-x^(2) * sin(x)(cos(x))^2 *(x*(5*(sin(x))^2-2)-3(sin(x)*cos(x))

And again :)
It is meant to be -3(*sin(x)*cos(x))

Answer 1

Sorry but there is nothing correct in your answer at all. It would score 0 marks both because it is wrong and because you haven't used the specified method.

lny = ln(x³sin²xcos³x) would be the start

lny = ln x³ + lnsin²x + lncos³x

lny = 3lnx + 2lnsinx + 3lncosx

1/y. dy/dx = 3/x + 2cosx/sinx - 3sinx/cosx

dy/dx = y(3/x + 2cotx - 3tanx)

........ = x³sin²xcos³x(3/x + 2cotx - 3tanx)

unless your question was y = x³sin(x²)cos(x³)) but the method is the same for this

Answer 2

y = (x^3) * (sin(x)^2) * (cos(x)^3)
By your approach
y' = - 3 * (x^3) * (sin(x)^3) * (cos(x)^2) + 2 * (x^3) * sin(x) * (cos(x)^4) + 3 * (x^2) * (sin(x)^2) * (cos(x)^3)
y' = (x^2) * sin(x) * (cos(x))^2 * (- 3 * x * (sin(x))^2 + 2 * x * (cos(x))^2 + 3 * sin(x) * cos(x))
y' = (x^3) * (sin(x))^2 * (cos(x))^3 * (3/x + 2cos(x) / sin(x) - 3sin(x) / cos(x))
which matches your teacher's result, but not the method.

y' = x^2sin(x)(cos^2(x)(- 3xsin^2(x) + 2xcos^2(x) + 3sin(x)cos(x))
y' = x^2sin(x)(cos^2(x)( - 3xsin^2(x) - 2xsin^2(x) + 2xsin^2(x) + 2xcos^2(x) + 3sin(x)cos(x))
y' = x^2sin(x)(cos^2(x)( - 5xsin^2(x) + 2x(sin^2(x) + cos^2(x)) + 3sin(x)cos(x))
y' = x^2sin(x)(cos^2(x)( - 5xsin^2(x) + 2x + 3sin(x)cos(x))
y' = x^2sin(x)(cos^2(x)( -x(5sin^2(x) - 2) + 3sin(x)cos(x))
y' = - x^2sin(x)(cos^2(x)( x(5sin^2(x) - 2) - 3sin(x)cos(x))
which is your form which, although correct, matches neither your teacher's form nor method. At best you might expect 1 mark for correctness and originality.

The purpose of the "logarithmic differentiation" was to arrive at a much simpler problem by taking the log of the equation:
y = (x^3) * (sin(x)^2) * (cos(x)^3)
ln(y) = ln(x^3) + ln(sin(x)^2) + ln(cos(x)^3)
which is now a sum instead of a product.
ln(y) = 3ln(x) + 2ln(sin(x)) + 3ln(cos(x))
and the differentiation is much simpler:
(1/y)(dy/dx) = 3/x + 2cos(x)/sin(x) - 3sin(x)/cos(x)
(dy/dx) = y(3/x + 2cos(x)/sin(x) - 3sin(x)/cos(x))
y' = x^3sin^2(x)cos^3(x)(3/x + 2cos(x)/sin(x) - 3sin(x)/cos(x))
a much, much quicker way to arrive at the derivitave, with less chance for error.

Answer 3

Well, it does not seem you used log. diff well enough....you answer should be just like the teacher's.

I say..2 marks