calculus answer check

[cowgiljl]

Am I goong about these two problems correctly?
A) F(x) = (xsinx)(cosx) find F'(x)
f(x)*d/dx(g(x) + g(x) * d/dx F(x) product rule

F'(x) = xsinx d/dx (cosx) + cos x d/dx xsinx
= xsinx*(-sinx) + cos x *(xcosx)
= -xsin^2 x^2 + xcos^2 x^2 is that right

B) F(x) = xe^x/ x^2 + 1 find F'(x)
using the quotent rult formula
F'x) = [U]xe^x* d/dx (x^2 +1) - (x^2)* d/dx xe^x
(x^2+1)^2
xe^x(2x)-(x^2 + 1)(1e^x) / (x^2+1)^2
= (2x^2e^x) - (x^2e^x + e^x) / (x^2+1)^2
= x^2 + e^x / (x^2 +1)^2 is that right?

thank you for your help
joe

[photon_mass]

no
you are right about using the product rule.
F = (xsinx)(cosx)= (f(x)*g(x)) = pq
F' = f'(x)g(x) + f(x)g'(x) =p'q +q'p

where you are having trouble:
notice that p is a product of x and sinx, so when you find p' you need to use the product rule there as well.

for the quotient rule: p is the numerator, q is the denominator.
(p'q-q'p)/q^2

remember: if either p or q is itself a product or a quotient or a chain, you need to use the appropriate rule to find p' and q'

trick to make life (or at least derivatives of quotients) easier:
if you have to find F' of F=1/x you would use the quotient rule. to make it easier, notice that F=1/x =x^-1 . now you can use the power rule. if the denominator in F were something like F = (x+1)/(x-1)
that is the same as F = (x+1)(x-1)^-1 and now you would use the product rule. just remember that for q you need to use the chain rule. the quotient rule has just become useless! one less thing to remember for exams! yay!