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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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.
$R = \tau z$
$\dfrac{dR}{d\tau} = \dfrac{d}{d\tau}\left(\tau z \right)$
By the product rule, the right hand side becomes:
$z\dfrac{d}{d\tau}(\tau) + \tau \dfrac{d}{d\tau}(z) = 1\cdot z + \tau \dfrac{dz}{d\tau} = z+\tau \dfrac{dz}{d\tau}$
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.