Am I Solving These Calculus Derivative Problems Correctly?

[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!

 

[M98Ranger]

A) Your solution for F'(x) is correct. You correctly applied the product rule and simplified the expression.

B) Your solution for F'(x) is also correct. You used the quotient rule correctly and simplified the expression. Good job!