Differentiating cosmological redshift wrt time confusion

I'm trying to understand Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the universe by Davis and Lineweaver. On page 19 they give an equation (number 23) for redshift $z$:$$1+z=\frac{R_{0}}{R\left(t\right)},$$where $R_{0}$ is the scale factor at time of observation and $R\left(t\right)$ is the scale factor at time of emission. They then differentiate this with respect to time $t$ to get$$\frac{dt}{R\left(t\right)}=-\frac{dz}{R_{0}H\left(z\right)},$$where $H\left(z\right)$ is Hubble’s constant at the time an object with redshift, $z$, emitted the light we now see and where redshift is used instead of time to parametrize Hubble’s constant. I just cannot see how differentiating wrt time gives this equation. And where does the minus sign come from? Using my high school maths I'd say$$\frac{dz}{dt}=\frac{d\left(\frac{R_{0}}{R\left(t\right)},\right)}{dt},$$but where do I go from there? Thank you.

$$\frac{dz}{dt}=\frac{d\left(\frac{R_{0}}{R\left(t\right)},\right)}{dt} = R_0 \left(\frac{-1}{R(t)^2}\right) \frac{dR}{dt}$$ but by the definition of the Hubble parameter $$H(z) = \frac{1}{R}\frac{dR}{dt},$$ $$\frac{dz}{dt}= -R_0\frac{H(z)}{R(t)}$$ and $$\frac{dt}{R(t)} = - \frac{dz}{R_0 H(z)}$$
• Ahhh, I believe that is called implicit differentiation, which I'm only vaguely familiar with. So I say $u=1/R$ and using the chain rule $\frac{du}{dt}=\frac{du}{dR}\frac{dR}{dt}$? Is that correct? – Peter4075 May 10 '18 at 7:52