Worked examples · Sheet 3

Partial derivatives and gradient

Theory used: Partial derivatives Gradient

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Compute the partial derivatives of $f(x, y) = 3x^2 y - 2xy^3 + 5x$ and the gradient at the point $(1, -1)$ .

f_x(x, y) = 6xy - 2y^3 + 5, \qquad f_y(x, y) = 3x^2 - 6xy^2

f_x(1, -1) = -6 + 2 + 5 = 1, \qquad f_y(1, -1) = 3 - 6 = -3

\nabla f(1, -1) = (1, -3)

In computing $f_x$ the variable $y$ is treated as a constant. The term $3x^2 y$ is the constant $3y$ times $x^2$ , so its derivative is $6xy$ ; the term $-2xy^3$ is the constant $-2y^3$ times $x$ , so its derivative is $-2y^3$ ; the term $5x$ has derivative $5$ . In computing $f_y$ it is $x$ that is treated as a constant. The term $3x^2 y$ has derivative $3x^2$ , the term $-2xy^3$ has derivative $-6xy^2$ , and the term $5x$ , which does not contain $y$ , has derivative $0$ .

Then one substitutes $(x, y) = (1, -1)$ , taking into account that $(-1)^3 = -1$ and $(-1)^2 = 1$ . The gradient is the vector of the two partial derivatives, $(1, -3)$ .

Exercise 2

Compute the partial derivatives of $f(x, y) = x\,e^{2y} + \ln(x + y^2)$ and the gradient at the point $(1, 0)$ .

f_x(x, y) = e^{2y} + \frac{1}{x + y^2}, \qquad f_y(x, y) = 2x\,e^{2y} + \frac{2y}{x + y^2}

f_x(1, 0) = 1 + 1 = 2, \qquad f_y(1, 0) = 2 + 0 = 2

\nabla f(1, 0) = (2, 2)

The function is defined where $x + y^2 > 0$ , and the point $(1, 0)$ satisfies the condition, because $1 > 0$ . With respect to $x$ , the term $x\,e^{2y}$ is $x$ multiplied by the constant $e^{2y}$ , so its derivative is $e^{2y}$ . The logarithm is differentiated with the rule for composite functions, according to which the derivative of $\ln u$ is $\dfrac{u'}{u}$ , and the derivative of $u = x + y^2$ with respect to $x$ is $1$ .

With respect to $y$ , the factor $x$ is constant and the derivative of $e^{2y}$ is $2e^{2y}$ , so the first term gives $2x\,e^{2y}$ ; the derivative of $x + y^2$ with respect to $y$ is $2y$ , so the second gives $\dfrac{2y}{x + y^2}$ . At the point $(1, 0)$ we have $e^{0} = 1$ and $x + y^2 = 1$ , and the gradient is $(2, 2)$ .

Exercise 3

Compute the partial derivatives of $f(x, y) = \dfrac{x - y}{x^2 + y^2}$ and the gradient at the point $(1, 1)$ .

\begin{aligned} f_x(x, y) &= \frac{1 \cdot (x^2 + y^2) - (x - y) \cdot 2x}{(x^2 + y^2)^2} \\[6pt] &= \frac{-x^2 + 2xy + y^2}{(x^2 + y^2)^2} \end{aligned}

\begin{aligned} f_y(x, y) &= \frac{(-1) \cdot (x^2 + y^2) - (x - y) \cdot 2y}{(x^2 + y^2)^2} \\[6pt] &= \frac{-x^2 - 2xy + y^2}{(x^2 + y^2)^2} \end{aligned}

f_x(1, 1) = \frac{-1 + 2 + 1}{4} = \frac{1}{2}, \qquad f_y(1, 1) = \frac{-1 - 2 + 1}{4} = -\frac{1}{2}

\nabla f(1, 1) = \left( \frac{1}{2}, -\frac{1}{2} \right)

The function is defined outside the origin, where the denominator vanishes. It is a quotient, so one applies the quotient rule, once with respect to $x$ and once with respect to $y$ . With respect to $x$ the derivative of the numerator $x - y$ is $1$ and that of the denominator $x^2 + y^2$ is $2x$ . Expanding the numerator gives $x^2 + y^2 - 2x^2 + 2xy = -x^2 + 2xy + y^2$ .

With respect to $y$ the derivative of the numerator is $-1$ and that of the denominator is $2y$ . The numerator becomes $-x^2 - y^2 - 2xy + 2y^2 = -x^2 - 2xy + y^2$ . At the point $(1, 1)$ the denominator equals $(1 + 1)^2 = 4$ , and one obtains the gradient $\left( \dfrac{1}{2}, -\dfrac{1}{2} \right)$ .

Exercise 4

Compute the partial derivatives of $f(x, y) = x^{y}$ , defined for $x > 0$ , and the gradient at the point $(2, 3)$ .

f_x(x, y) = y\,x^{y - 1}, \qquad f_y(x, y) = x^{y} \ln x

f_x(2, 3) = 3 \cdot 2^{2} = 12, \qquad f_y(2, 3) = 2^{3} \ln 2 = 8 \ln 2

\nabla f(2, 3) = (12, \ 8 \ln 2)

The two partial derivatives require two different formulas, because the variable appears once in the base and once in the exponent. In computing $f_x$ the exponent $y$ is a constant, so $x^{y}$ is a power and its derivative is $y\,x^{y - 1}$ . In computing $f_y$ the base $x$ is a positive constant, so $x^{y}$ is an exponential with base $x$ and its derivative is $x^{y} \ln x$ , according to the table of derivatives of elementary functions.

At the point $(2, 3)$ one obtains $f_x = 3 \cdot 4 = 12$ and $f_y = 8 \ln 2 \approx 5.55$ . Swapping the two formulas is a frequent error, and it is avoided by establishing, before differentiating, which of the two letters is held fixed.