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Thread: Calculus: what rules are being use here?

  1. #1
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    Calculus: what rules are being use here?

    Hi,

    Please see the picture attached. I just don't understand the move from 1 to 2. Is it some combination of rules?

    Thanks

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  2. #2
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    Re: Calculus: what rules are being use here?

    You are given that the function $z$ is a function of the variable $1-\tau$. So, the derivative of $R$ with respect to $\tau$ is:

    $\dfrac{dR}{d\tau} = z+\tau \dfrac{dz}{d\tau}$

    Now, $\dfrac{dz}{d\tau} = \dfrac{dz}{d(1-\tau)}\dfrac{d(1-\tau)}{d\tau} = -\dfrac{dz}{d(1-\tau)}$ by the Chain rule.
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    Re: Calculus: what rules are being use here?

    Quote Originally Posted by SlipEternal View Post
    You are given that the function $z$ is a function of the variable $1-\tau$. So, the derivative of $R$ with respect to $\tau$ is:

    $\dfrac{dR}{d\tau} = z+\tau \dfrac{dz}{d\tau}$

    Now, $\dfrac{dz}{d\tau} = \dfrac{dz}{d(1-\tau)}\dfrac{d(1-\tau)}{d\tau} = -\dfrac{dz}{d(1-\tau)}$ by the Chain rule.
    Thanks. Are you able to break that first step down for me anymore? I'm not that smart.

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  4. #4
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    Re: Calculus: what rules are being use here?

    $R = \tau z$

    $\dfrac{dR}{d\tau} = \dfrac{d}{d\tau}\left(\tau z \right)$

    By the product rule, the right hand side becomes:

    $z\dfrac{d}{d\tau}(\tau) + \tau \dfrac{d}{d\tau}(z) = 1\cdot z + \tau \dfrac{dz}{d\tau} = z+\tau \dfrac{dz}{d\tau}$
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    Re: Calculus: what rules are being use here?

    Ah R = tz! I thought R = tz (1 -t). What a silly mistake thank you so much!

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  6. #6
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    Re: Calculus: what rules are being use here?

    Quote Originally Posted by tonytango View Post
    Ah R = tz! I thought R = tz (1 -t). What a silly mistake thank you so much!

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    I see your confusion. You are used to seeing functions represented by the letter $f $. So, $f (x) $ is the function evaluated at $x $. In this case, $z (1-\tau)$ is the function $z $ evaluated at $1-\tau $. That's what was meant by the phrase $z $ is a function of $1-\tau $ in the original image you posted.
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    Re: Calculus: what rules are being use here?

    Quote Originally Posted by SlipEternal View Post
    You are given that the function $z$ is a function of the variable $1-\tau$. So, the derivative of $R$ with respect to $\tau$ is:

    $\dfrac{dR}{d\tau} = z+\tau \dfrac{dz}{d\tau}$

    Now, $\dfrac{dz}{d\tau} = \dfrac{dz}{d(1-\tau)}\dfrac{d(1-\tau)}{d\tau} = -\dfrac{dz}{d(1-\tau)}$ by the Chain rule.
    Sorry to bother you again, but could you explain how to move to the last part here? -\dfrac{dz}{d(1-\tau)}$

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