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Thread: chain rule problem, with 2 constants!

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

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

    How do I go about this?

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

    Is this is correct?

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

    OR this one is the correct one?

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

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

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

    $\displaystyle \frac{d}{dx}(k\cdot f(u))=k\cdot\frac{d}{dx}(f(u))$ where $\displaystyle k$ 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

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

    The text book has

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

    an extra $\displaystyle \pi$ 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 $\displaystyle \pi$ comes from that you are missing.
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    Re: chain rule problem, with 2 constants!

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

    then

    $\displaystyle -\sin \pi w$

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

    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:

    $\displaystyle \frac{d}{dw}(\cos(\pi w))=-\sin(\pi w)\cdot\frac{d}{dw}(\pi w)=-\pi\sin(\pi w)$
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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:

    $\displaystyle \frac{d}{dw}(\cos(\pi w))=-\sin(\pi w)\cdot\frac{d}{dw}(\pi w)=-\pi\sin(\pi w)$
    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 $\displaystyle \pi 5$

    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:

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

    If we let:

    $\displaystyle f_1(x)=x^2$

    $\displaystyle f_2(x)=\cos(x)$

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

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

    The chain rule tells us then:

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