Thread: Calculus: what rules are being use here?

1. 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. 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.

3. Re: Calculus: what rules are being use here?

Originally Posted by SlipEternal
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. 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:

$z\dfrac{d}{d\tau}(\tau) + \tau \dfrac{d}{d\tau}(z) = 1\cdot z + \tau \dfrac{dz}{d\tau} = z+\tau \dfrac{dz}{d\tau}$

5. 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. Re: Calculus: what rules are being use here?

Originally Posted by tonytango
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.

7. Re: Calculus: what rules are being use here?

Originally Posted by SlipEternal
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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