Product and Quotient rules

Question

I am trying to assist my daughter doing her college calculus homework, but it;s over my head. Here is the problem:

If f(x)=2xsin(x)cos(x), find f(x)

Also, find f(3)

Thanks in advance.

Answer 1

f(x) =2xsinxcosx

= xsin(2x)

f'(x) = xcos(2x)(2) + sin(2x)

= 2xcos(2x) + sin(2x)

(using the identity 2sinxcos = sin(2x))

notice sin^2(x) = (sinx)^2 and sin(2x) is very different. Latter doubles the angle x to 2x

former keeps the angle x the same and squares sinx = (sinx)^2, written as sin^2(x)

you could, if you wanted, revert back to

= 2xcos^2(x) - 2xsin^2(x) + 2sinxcosx

or alternatively

f(x) = 2xsinxcosx

f'(x) = 2x(sinx(-sinx) + cosx(cosx)) + 2sinxcosx

= 2x(cos^2(x) - sin^2(x)) + 2sinxcox

= 2xcos(2x) +sin(2x)

IF you come up with the same answer working it 2 different ways, odds are good it's the right answer, in one form or another

two identities that come up are

sin(2x) = 2sinxcosx

and

cos(2x) = cos^2(x) + sin^2(x)

1st method was do the derivative of a product twice,

where the product is the multiplication of 2x times sinx times cosx

treat it as 2 factors 2x and sinxcosx, then when you take the derivative of sinxcosx apply the product rule again

alternative method is avoid that

by changing 2sinxcosx into sin(2x)

then you only have to deal with x and sin(2x) applying the product rule just once

to get 1st term times derivative of 2nd term Plus 2nd term times derivative of 1st term

to get

x(cos(2x)(2) + sin(2x)

= 2x(cos(2x) + sin(2x)

work it either way, you end up with

f'(x) = 2x(cos(2x)) + sin(2x)

if you want to find f(3) or f'(3), substitute 3 for x

but becareful, as it's not clear from what's given when 3 = 3 degrees or if it equals 3 radians

the answer is very different depending on which it is

3 degrees is very small, close to zero. 3 radians is very large, close to 180 degrees

sin3 degrees = about .05

sin 3 radians = about sin 171.89 = about .141

cos3 degrees = about .999

cos 3radians = about .990

f(3 degrees)= 2(3)sin3cos3 = 6(.05)(.999) = about 0.3

f(3 radians) = 2(3)(.141)(.99) = 0.838

f'(x) = 2xcos(2x) + sin(2x)

f'(3 degrees) = 2(3)cos6 + sin6 = 6.072

f'(3 radians) = 2(3)cos(6(180/pi)) + sin(6(180)/pi = about 5.482

you have to use a calculator, some are set using degrees, some on radians

if you mix them up, you get very different answers