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Math Help - chain rule problem, with 2 constants!

  1. #1
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    chain rule problem, with 2 constants!

    \frac{d}{dw}{[a\cos^{2}\pi w +b\sin^{2}\pi w ]} (a,b constants)

    How do I go about this?

    The constant "a" or "b" independent from cos/sin here?..

    Is this is correct?

    \frac{d}{dw}{[(a\cos\pi w)^{2} +(b\sin\pi w)^{2} ]}

    OR this one is the correct one?

    \frac{d}{dw}{[a(\cos\pi w)^{2} +b(\sin\pi w)^{2} ]}
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  2. #2
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    Re: chain rule problem, with 2 constants!

    The second one is equivalent to the original statement.
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    Re: chain rule problem, with 2 constants!

    then we have to use the product rule right?
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  4. #4
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    Re: chain rule problem, with 2 constants!

    No, you have a sum, not a product. Differentiating term by term, I would use the power and chain rules.
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    Re: chain rule problem, with 2 constants!

    Quote Originally Posted by MarkFL2 View Post
    No, you have a sum, not a product. Differentiating term by term, I would use the power and chain rules.
    Right sum but isn't there a product rule too here?
    \frac{d}{dw}{[a(\cos\pi w)^{2} ]}

    a \and \(\cos\pi w)^{2}
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  6. #6
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    Re: chain rule problem, with 2 constants!

    If a and b are constants, while you could technically use the product rule, I would simply use the rule:

    \frac{d}{dx}(k\cdot f(u))=k\cdot\frac{d}{dx}(f(u)) where k is an arbitrary constant.

    You will get the same result using the product rule since the derivative of a constant is zero.
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    Re: chain rule problem, with 2 constants!

    I solved the problem but my result is differing from the text book..

    I got

    {-2a\cos\pi w \sin\pi w + 2b \sin\pi w \cos\pi w}

    The text book has

    {-2\pi a\cos\pi w \sin\pi w + 2\pi b \sin\pi w \cos\pi w}

    an extra  \pi between 2a and 2b.
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: chain rule problem, with 2 constants!

    You have only partially applied the chain rule. You must also apply it to the arguments of the trig. functions. This is where the \pi comes from that you are missing.
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    Re: chain rule problem, with 2 constants!

    this is what I did
    \frac{d}{dw}(\cos \pi w)

    then

    -\sin \pi w

    I should have to applied product rule here? or maybe rewrite \frac{d}{dw}(\cos \pi w) to \frac{d}{dw}(\cos \pi) \frac{d}{dw}(\cos w)

    is that is my mistake?
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  10. #10
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    Re: chain rule problem, with 2 constants!

    You need in the case from your previous post:

    \frac{d}{dw}(\cos(\pi w))=-\sin(\pi w)\cdot\frac{d}{dw}(\pi w)=-\pi\sin(\pi w)
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  11. #11
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    Re: chain rule problem, with 2 constants!

    Quote Originally Posted by MarkFL2 View Post
    You need in the case from your previous post:

    \frac{d}{dw}(\cos(\pi w))=-\sin(\pi w)\cdot\frac{d}{dw}(\pi w)=-\pi\sin(\pi w)
    I find it confusing to decide how deep I should use the change rule. What Should I know.. how do I decide..

    if let say some constant 5 for example was here instead of w.. so it would have looked like \pi 5

    what could been the scenario?
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  12. #12
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    Re: chain rule problem, with 2 constants!

    A trig. function of a constant is just a constant, so its derivative would be zero.

    How deeply you must go is determined by how many "compositions" you have.

    The two terms you began with have 3 compositions each. Let's look at the first term:

    a\cos^2(\pi w)=a(\cos(\pi w))^2

    If we let:

    f_1(x)=x^2

    f_2(x)=\cos(x)

    f_3(x)=\pi x then we may state:

    a(\cos(\pi w))^2=a\cdot f_1(f_2(f_3(w)))

    The chain rule tells us then:

    \frac{d}{dw}(a\cdot f_1(f_2(f_3(w))))=a\cdot f_1'(f_2(f_3(w)))\cdot f_2'(f_3(w))\cdot f_3'(w)
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