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Calculus answer check

  1. Sep 29, 2004 #1
    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
     
  2. jcsd
  3. Sep 29, 2004 #2
    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!
     
    Last edited: Sep 29, 2004
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