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Math Help - Composite Rule

  1. #1
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    Composite Rule

    Hi can some please check my working out and answer?

    Use the composite rule to differentiate the function
    f(x)=e^x(2/3sin x)

    My answer:
    e^x(2/3sinx)

    2/3 sin x = 2sin x/3

    e^x(2sin x/3)

    2e^xsin x/3

    Thanks in advance
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Well it's not clear what the expression is. Use LaTeX. If the expression is what I think it is, then you haven't done anything at all to it.
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  3. #3
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    Quote Originally Posted by stewpot View Post
    Hi can some please check my working out and answer?

    Use the composite rule to differentiate the function
    f(x)=e^x(2/3sin x)

    My answer:
    e^x(2/3sinx)

    2/3 sin x = 2sin x/3

    e^x(2sin x/3)

    2e^xsin x/3

    Thanks in advance
    It would help if you used more parentheses. The "composite" rule, I would think of as involving composite functions: f(g(x)) but what you have written appears to be (e^x) ((2/3)sin(x)) rather than the composite e^((2/3)x sin(x)). In any case, you don't seem to differentiated at all! You final "answer", 2(e^x)(sin(x))/3, is just the original function. What is the derivative of that function?
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    It would help if you used more parentheses. The "composite" rule, I would think of as involving composite functions: f(g(x)) but what you have written appears to be (e^x) ((2/3)sin(x)) rather than the composite e^((2/3)x sin(x)). In any case, you don't seem to differentiated at all! You final "answer", 2(e^x)(sin(x))/3, is just the original function. What is the derivative of that function?
    Hi yep your right I hadn't done anything! Had another look could you cast your eye over it?
    f(x)=e^x(2/3sin x)

    f(x)=e^x(2/3sin(x) (2/3sin(x))

    =e^x(2/3sin(x) 2/3cos(x)

    =2/3cos(x) e^x(2/3sin(x))
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  5. #5
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    Quote Originally Posted by stewpot View Post
    Hi yep your right I hadn't done anything! Had another look could you cast your eye over it?
    f(x)=e^x(2/3sin x)
    This now says f(x)= e^x (\frac{2}{3}sin(x)) is that what you intended?

    f(x)=e^x(2/3sin(x) (2/3sin(x))
    I have no idea what you mean by this! it is certainly NOT the function above. Did you mean that as f'(x)? It is not for either f(x)= e^x(\frac{2}{3}sin(x)) or f(x)= e^{\frac{2}{3}x sin(x)}.

    =e^x(2/3sin(x) 2/3cos(x)

    =2/3cos(x) e^x(2/3sin(x))
    If you meant f(x)= e^{\frac{2}{3}x sin(x)}, its derivative is itself, e^{\frac{2}{3}x sin(x)} times the derivative of \frac{2}{3}x sin(x) which is NOT \frac{2}{3}cos(x).
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  6. #6
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    Question Got the same problem!

    I'm stuck on trying to differentiate f(x)=e^x(2/3sin x) as well and I've tangled myself up into a complete mind block! I think it's the 2/3 aspect confusing me the most, I've never been quite sure what to do with fractions when they pop up like this in functions. D'oh. Any help/advice would be much appreciated!
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  7. #7
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    If this really is f(x)= e^x((2/3)sin(x)), then the "2/3", a constant, just multiplies the whole thing. The derivative is f'(x)= (2/3)(e^x sin(x)+ e^x cos(x))= (2/3)e^x(sin(x)+ cos(x))

    If it is actually the composite function, as suggested by the title, f(x)= e^{(2/3)x sin(x)} the, by the composite rule (also called the "chain rule") f'(x)= e^{(2/3)x sin(x)}\left((2/3)x sin(x)\right)'= 2/3(sin(x)+ xcos(x))e^{(2/3)x sin(x)}
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  8. #8
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    composite rule

    I am still unsure as to whether this formula has been entered correctly. The correct formula is in the attached file.

    Can you point me in the right direction please?
    Attached Files Attached Files
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  9. #9
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    the rule is \frac{d}{dx} e^u = e^u \cdot \frac{du}{dx} , where u is a function of x.

    for your problem ...

    y = e^{\frac{2}{3}\sin{x}}

    \frac{dy}{dx} = e^{\frac{2}{3}\sin{x}} \cdot \frac{2}{3}\cos{x}
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