Math · Calculus I · Worked example

Differentiate a quotient of polynomials

Differentiate f(x) = (x² + 1)/(x − 3), then find where its tangent line is horizontal.

f (x) = \frac{x^{2} + 1}{x - 3}

Name the top and the bottom

The top is x² + 1, with derivative 2x. The bottom is x − 3, with derivative 1. The function is undefined at x = 3.

top = x^{2} + 1, bottom = x - 3, x \neq 3

\begin{array}{c} top = x^{2} + 1 \\ bottom = x - 3, x \neq 3 \end{array}

Apply the quotient rule

Bottom times the derivative of the top, minus top times the derivative of the bottom, over the bottom squared.

f^{'} (x) = \frac{2 x (x - 3) - (x^{2} + 1) (1)}{{(x - 3)}^{2}}

Simplify the numerator

2x(x − 3) = 2x² − 6x. Subtracting x² + 1 leaves x² − 6x − 1. Keep the denominator as (x − 3)².

f^{'} (x) = \frac{x^{2} - 6 x - 1}{{(x - 3)}^{2}}

Find the horizontal tangents

The tangent is horizontal where f′(x) = 0, which happens where the numerator is zero and the denominator is not. x² − 6x − 1 does not factor over the integers, so use the quadratic formula: x = 3 ± √10. Neither value is 3, so both count.

\begin{array}{l} x^{2} - 6 x - 1 = 0 \\ x = 3 \pm \sqrt{10} \end{array}

Locate the points

x = 3 + √10 ≈ 6.162 gives f(x) = 6 + 2√10 ≈ 12.325, a low point of the right branch; x = 3 − √10 ≈ −0.162 gives f(x) = 6 − 2√10 ≈ −0.325, a high point of the left branch.

Result

f′(x) = (x² − 6x − 1)/(x − 3)² for x ≠ 3. The tangent is horizontal at x = 3 ± √10, about 6.162 and −0.162.

f^{'} (x) = \frac{x^{2} - 6 x - 1}{{(x - 3)}^{2}}

Your turn

Differentiate y = sin x / x.

Show the answer and explanation

dy/dx = (x cos x − sin x)/x², for x ≠ 0.

Bottom times the derivative of the top is x cos x; top times the derivative of the bottom is sin x · 1. So dy/dx = (x cos x − sin x)/x².

\begin{array}{l} y = \frac{sin x}{x} \\ \frac{d y}{d x} = \frac{x cos x - sin x}{x^{2}} \end{array}

Keep exploring

Graph f and look at x ≈ −0.162 and x ≈ 6.162: the curve levels off exactly where the numerator of f′ is zero.

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