Worked examples · Sheet 2

Derivatives of products and quotients

Theory used: Differentiation rules Derivatives of elementary functions

Exercise 1

Compute the derivative of $f (x) = (x^{3} - 2 x) e^{x}$ .

f^{'} (x) = (3 x^{2} - 2) e^{x} + (x^{3} - 2 x) e^{x} = (x^{3} + 3 x^{2} - 2 x - 2) e^{x}

The function is the product of a polynomial and an exponential, so the product rule applies: derivative of the first factor times the second, plus the first factor times the derivative of the second. The derivative of $x^{3} - 2 x$ is $3 x^{2} - 2$ and that of $e^{x}$ is $e^{x}$ itself. Factoring out $e^{x}$ , which is always positive, the sign of the derivative is the sign of the polynomial $x^{3} + 3 x^{2} - 2 x - 2$ ; the factored form is the one that is useful in curve sketching.

Exercise 2

Compute the derivative of $f (x) = \frac{x ^{2} + 1}{x - 3}$ .

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

For a quotient one uses the quotient rule: in the numerator, the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, and in the denominator, the square of the denominator. The order of the two terms in the numerator matters, because swapping them changes the sign. Expanding and combining like terms gives $x^{2} - 6 x - 1$ . The denominator $(x - 3)^{2}$ is left in the form of a square; since it is positive for $x \neq = 3$ , the sign of $f^{'}$ is that of the numerator.

Exercise 3

Compute the derivative of $f (x) = x ln x$ and find the point where the tangent to the graph is horizontal.

f^{'} (x) = 1 \cdot ln x + x \cdot \frac{1}{x} = ln x + 1, x > 0

f^{'} (x) = 0 ⟺ ln x = - 1 ⟺ x = e^{- 1} = \frac{1}{e}

f (\frac{1}{e}) = \frac{1}{e} \cdot (- 1) = - \frac{1}{e}

y = x \ln x

The graph of

x ln x

has a horizontal tangent at the point

(\frac{1}{e}, - \frac{1}{e})

The function is defined for $x > 0$ , where the logarithm exists. With the product rule the first factor $x$ has derivative $1$ , and the second, $ln x$ , has derivative $\frac{1}{x}$ , which multiplied by $x$ gives $1$ . The tangent is horizontal where the slope is zero, that is, where $f^{'} (x) = 0$ ; the equation $ln x = - 1$ is solved by applying the exponential to both sides. The required point is $(\frac{1}{e}, - \frac{1}{e})$ , that is, approximately $(0.368, - 0.368)$ .