An alternative way to the quotient rule

The quotient rule helps you to find the derivative of a function that consists of two functions divided by each other. We can prove it from first principles, but there’s another way to arrive at the rule by implicite differentiation. It goes like this.

Let’s look at the function

    \[y(x)=\frac{f(x)}{g(x)}\]

Now we take the natural logarithm (but it could be any logarithm) of both sides.

    \[\ln(y(x))=\ln(\frac{f(x)}{g(x)})\]

As per the rules of logarithms, we know that

    \[\ln(a)-\ln(b)=\ln(\frac{a}{b})\]

so we can write

    \[\ln(y(x))=\ln(f(x))-\ln(g(x))\]

Now let’s find the derivative of \ln(y(x)) using implicite differentiation and the chain rule:

    \[\frac{1}{y(x)}\cdot y'(x)=\frac{1}{f(x)}\cdot f'(x)-\frac{1}{g(x)}\cdot g'(x)\]

or

    \[\frac{y'(x)}{y(x)}=\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}\]

Now let’s create equal denominators:

    \[\frac{y'(x)}{y(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{f(x)g(x)}\]

Multiply both sides by y(x) and we get

    \[y'(x)=y(x)\cdot\frac{f'(x)g(x)-f(x)g'(x)}{f(x)g(x)}\]

Remember that we know

    \[y(x)=\frac{f(x)}{g(x)}\]

so

    \[y'(x)=\frac{f(x)}{g(x)}\cdot\frac{f(x)'g(x)-f(x)g'(x)}{f(x)g(x)}\]

Eliminating f(x) in the numerator and denominator brings us at

    \[y'(x)=\frac{1}{g(x)}\cdot\frac{f'(x)g(x)-f(x)g'(x)}{g(x)}\]

and we’re done:

    \[y'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}\]

It’s not much, but for those in need it might be just enough. 🙂