# How do I calculate the transfer function of this basic terminated RC filter?

I'm reading through an electronics book to teach myself, and I'm in the section about filters. I've been following along so far, but now I'm confused about how the author came to a conclusion in his math. Here is an excerpt from the book (Practical Electronics for Inventors, 4th edition, page 213): He is computing the transfer function for the circuit in the image, and he says that: $$H=\frac{V_{out}}{V_{in}}=\frac{1/(j\omega C)||R_L}{R+[1/(j\omega C)||R_L]}$$ This makes sense to me, since this is just computing $$\V_{out}\$$ using the voltage divider equation. The next step is what confuses me, where he says that the equation above is equivalent to: $$= \frac{R'}{1+j(\omega R'C)}V_{in} \text{ where } R'=R||R_L$$

How do I come to this conclusion? I've tried simplifying the circuit using Thevenin's theorem by combining $$\R\$$ and $$\C\$$, which gives me: $$R_{THEV} = R||\frac{1}{j\omega C} = \frac{R/(j\omega C)}{R+1/(j\omega C)} = \frac{R}{1+j\omega RC}$$

That's pretty close to the author's answer, except I have $$\R\$$ instead of $$\R'\$$. I've searched around and haven't found other resources making this leap. I'm a little lost! The book continues making use of this throughout the filters section, so I really want to understand it. Any help here would be greatly appreciated.

Edit: The answer by Paul below solved the problem. The numerator of the equation in red should be $$\R'/R\$$ and not $$\R\$$. I expanded both the green equation and the corrected red equation and got a matching result.

Expansion of green equation: $$H=\frac{V_{out}}{V_{in}}=\frac{1/(j\omega C)||R_L}{R+[1/(j\omega C)||R_L]}V_{in} = \frac{(R_L/(j\omega C))/(R_L + 1/(j\omega C))}{R+(R_L/(j\omega C))/(R_L + 1/(j\omega C))}\\ = \frac{R_L/(R_L j\omega C+1)}{R+R_L/(R_L j\omega C+1)} = \frac{R_L}{R(R_L j\omega C+1)+R_L} = \frac{R_L}{R+R_L+j\omega CRR_L}$$

Expansion of corrected red equation (with $$\R'\$$ replaced with $$\R||R_L\$$): $$\frac{(R||R_L)/R}{1+j\omega (R||R_L)C} = \frac{\frac{RR_L}{R+R_L}/R}{1+j\omega C(\frac{RR_L}{R+R_L})} = \frac{R_L}{(R+R_L)(1+j\omega C\frac{RR_L}{R+R_L})} = \frac{R_L}{R+R_L+j\omega CRR_L}$$