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Differentiate the function

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    The problem is to differentiate the function.
    y=cos4(x2 + ex)


    2. Relevant equations
    cos(x)'= -sin(x), (xn)'=n*xn-1, (ex)'=ex*x'


    3. The attempt at a solution
    Thought it would be the chain rule. Here's what I came up with:

    y'=4cos3(-sin(x2 + ex)(2x + ex)

    Is this right? If not, any suggestions on what I did wrong?
     
    Last edited: Dec 8, 2008
  2. jcsd
  3. Dec 8, 2008 #2

    gabbagabbahey

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    Is this a typo? What are you taking the cosine of?
     
  4. Dec 8, 2008 #3
    no it's not a typo. my answer was y'= 4cos3(-sin(x2 + ex))*(2x + ex)
     
  5. Dec 8, 2008 #4
    I did forget to say that the problem says not to simplify.
     
  6. Dec 8, 2008 #5
    You're close but take another look at the chain rule.

    f(x) = h(g(x))

    f'(x) = h'(g(x)) * g'(x)

    Your answer follows the following incorrect differentiation.

    f'(x) = h'(g'(x))

    But you only messed up for one part of the chain. The innermost function was differentiated correctly.
     
  7. Dec 8, 2008 #6

    gabbagabbahey

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    In that case it is incorrect.

    Try using the substitution [itex]u=x^2+e^x[/itex] along with the chain rule....What is [tex]\frac{d}{dx}\cos^4(u)[/tex]?
     
  8. Dec 8, 2008 #7
    so h(x)=cos4(x2+ex)

    and g(x)= x2 + ex

    but would the cos4 part be another function by itself? So then it would be f(g(h(x))).

    and by definition of the chain rule it would be f '(g(h(x)))*g'(h(x))*h'(x)?
     
  9. Dec 8, 2008 #8

    gabbagabbahey

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    hmmm this doesn't make much sense.

    Instead, let's call [itex]g(x) \equiv \cos^4(x)[/itex] and [itex]h(x)\equiv x^2+e^x[/itex] then [itex]f(x)\equiv\cos^4(x^2+e^x)=g(h(x))[/itex]

    ....follow?

    Now, what does the chain rule give you for [itex]f'(x)[/itex]?
     
  10. Dec 8, 2008 #9
    yeah, i think i follow. so from your definition, f'(x)=g'(h(x))*h'(x)?
     
  11. Dec 8, 2008 #10
    so, (cos(x^2+e^x))^4

    f'(x)= 4cos(x^2+e^x)^3*-sin(x^2+e^x)*(2x+e^x)
     
  12. Dec 8, 2008 #11

    gabbagabbahey

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    Yes, much better!:approve:
     
  13. Dec 8, 2008 #12
    you guys are awesome
     
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