I need help on the process of solving this derivative.

Question

How do I go about solving this derivative. $$f(x)=\ln\left(\frac{7x}{x+4}\right)$$ I go from this to $$1. \quad f(x)=\ln(7)+\ln(x)-\ln(x+4)$$ and then $$2. \quad f'(x)=\frac{1}{x}-\frac{1}{x+4}$$ then $$3. \quad f'(x)=\frac{x+4-x}{x(x+4)}$$ and end up with $$4. \quad f'(x)=\frac{4}{x(x+4)}$$ which is the correct answer. This procedure is what we did in class, but I do not get how I went from step 2 to 3 which is what has me confused. A step by step explanation would be greatly appreciated. Thanks!

Answer 1

That was simply finding the common denominator, which is $x(x+4)$.

$x$ divides $x(x+4)$ to give us $1\cdot (x+4)$ in the numerator, and $x+4$ divides $x(x+4)$ to give us $1\cdot x$ which we subtract from the first term in the numerator. Then simplify $(x+4) - x = 4$ in the numerator.

Answer 2

Using the chain rule: $$f'(x)=\left(\frac{x+4}{7x}\right)\left(\frac{7x}{x+4}\right)'=\left(\frac{x+4}{7x}\right)\left(\frac{28}{(x+4)^2}\right)=\frac{4}{x(x+4)}$$