The derivative of a quotient and the tangent to a rational function

We find the derivative of $\displaystyle {f(x)}={\frac{1}{x}}$ by the definition, as we found the derivative of $\displaystyle x^2$. The working below puts the two fractions in the numerator over a common denominator:

• As $\displaystyle h\to0$, the denominator $\displaystyle x(x+h)$ approaches $\displaystyle {x\cdot x}={x^2}$.
• So the average rate of change approaches $\displaystyle -\frac{1}{x^2}$.

The derivative is negative throughout the domain: the function $\displaystyle \frac{1}{x}$ falls to the left of 0 and to the right of it too.

A quotient is a product by the reciprocal: $\displaystyle {\frac{f(x)}{g(x)}}={f(x)\cdot\frac{1}{g(x)}}$. The proof below differentiates the two factors by the rules we already have:

• $\displaystyle \frac{1}{g(x)}$ is a composite function: the outer function $\displaystyle \frac{1}{u}$, and the inner one $\displaystyle {u}={g(x)}$. So $\displaystyle {\left(\frac{1}{g(x)}\right)'}={-\frac{1}{[g(x)]^2}\cdot g'(x)}$.
• By the product rule:

• The order in the numerator matters: the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator.
• The denominator $\displaystyle [g(x)]^2$ is usually left as it is, brackets unopened.

The example below differentiates $\displaystyle {f(x)}={\frac{2x+1}{x-3}}$:

• The numerator is $\displaystyle 2x+1$, and its derivative is 2.
• The denominator is $\displaystyle x-3$, and its derivative is 1.

The numerator $\displaystyle -7$ is negative and the denominator positive, so $\displaystyle f'(x)<0$ throughout the domain.

The tangent is found as for a polynomial: $\displaystyle {y_1}={f(x_1)}$, $\displaystyle {m}={f'(x_1)}$, and then $\displaystyle {y-y_1}={m(x-x_1)}$.

The example below finds the tangent to the graph of $\displaystyle {f(x)}={\frac{2x+1}{x-3}}$ at the point where $\displaystyle {x}={4}$:

• $\displaystyle {f(4)}={\frac{9}{1}}={9}$: the point of contact is $\displaystyle (4,9)$.
• $\displaystyle {f'(4)}={\frac{-7}{1^2}}={-7}$: the slope of the tangent.
• $\displaystyle {y-9}={-7(x-4)}$, that is $\displaystyle {y}={-7x+37}$.

The function is $\displaystyle {f(x)}={\frac{1}{x+3}}$.
Its average rate of change from $\displaystyle x$ to $\displaystyle x+h$ is $\displaystyle \frac{\frac{1}{x+h+3}-\frac{1}{x+3}}{h}$.

Simplify the expression.

What does your expression approach as $\displaystyle h\to0$?
What is $\displaystyle f'(x)$?

Differentiate.

Here the numerator is a constant: you can use the quotient rule, or the rule for a composite function.

The function is $\displaystyle {f(x)}={\frac{x}{x-1}}$.

Find $\displaystyle f'(x)$.

Work out the slope of the tangent to the graph at the point $\displaystyle (2,2)$.

Write on the board the equation of the tangent at the point $\displaystyle (2,2)$.

The function is $\displaystyle {f(x)}={\frac{x+3}{x-1}}$.
Its derivative is $\displaystyle {f'(x)}={\frac{-4}{(x-1)^2}}$.

Find the $\displaystyle x$-coordinates of the points on the graph where the slope of the tangent is $\displaystyle -1$.

Write on the board the equations of the two tangents at these points.

The function is $\displaystyle {f(x)}={\frac{1}{x}}$.
At two points the tangents to the graph are parallel to the line $\displaystyle {y}={-4x}$.

$\displaystyle {f'(x)}={-\frac{1}{x^2}}$, so the $\displaystyle x$-coordinates of the points are the solutions of the equation $\displaystyle {-\frac{1}{x^2}}={-4}$.
Solve it.

Write the two points of contact and the equations of the two tangents.
How are the two points related?