Worked examples · Sheet 1

Computing partial derivatives and the gradient

Theory used: Partial derivative Gradient

Exercise 1

Compute the partial derivatives of

f (x, y) = x^{3} y^{2} - 2 x y + 5 y

and the gradient $\nabla f (1, 2)$ .

f_{x} (x, y) = 3 x^{2} y^{2} - 2 y, f_{y} (x, y) = 2 x^{3} y - 2 x + 5

f_{x} (1, 2) f_{y} (1, 2) = 3 \cdot 1 \cdot 4 - 2 \cdot 2 = 8 = 2 \cdot 1 \cdot 2 - 2 + 5 = 7

\nabla f (1, 2) = (8, 7)

To compute $f_{x}$ one differentiates with respect to $x$ treating $y$ as a constant: the term $x^{3} y^{2}$ has derivative $3 x^{2} y^{2}$ , the term $- 2 x y$ has derivative $- 2 y$ , and the term $5 y$ is constant with respect to $x$ , so its derivative is zero. For $f_{y}$ the roles are swapped: $x^{3} y^{2}$ gives $2 x^{3} y$ , $- 2 x y$ gives $- 2 x$ and $5 y$ gives $5$ . The gradient at $(1, 2)$ is obtained by substituting the coordinates of the point into the two derivatives, therefore $\nabla f (1, 2) = (8, 7)$ .

Exercise 2

Compute the gradient of

f (x, y) = (x^{2} + y) e^{x y}

at the point $(1, 0)$ .

f_{x} (x, y) = 2 x e^{x y} + (x^{2} + y) y e^{x y} = e^{x y} (2 x + x^{2} y + y^{2})

f_{y} (x, y) = e^{x y} + (x^{2} + y) x e^{x y} = e^{x y} (1 + x^{3} + x y)

f_{x} (1, 0) = e^{0} (2 + 0 + 0) = 2, f_{y} (1, 0) = e^{0} (1 + 1 + 0) = 2

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

The function is a product, so for each partial derivative one applies the product rule in the variable of differentiation. Differentiating with respect to $x$ , the factor $x^{2} + y$ has derivative $2 x$ and the factor $e^{x y}$ has derivative $y e^{x y}$ , because the exponent $x y$ has derivative $y$ with respect to $x$ . Differentiating with respect to $y$ , the first factor has derivative $1$ and the exponential has derivative $x e^{x y}$ . In both cases one factors out $e^{x y}$ to make the result compact. At the point $(1, 0)$ the exponential equals $e^{0} = 1$ , so the polynomials $2 x + x^{2} y + y^{2}$ and $1 + x^{3} + x y$ remain, and both equal $2$ .

Exercise 3

Compute the gradient of

f (x, y) = ln (x^{2} + y^{2}) + arctan \frac{y}{x}

at the point $(1, 1)$ , in the half-plane $x > 0$ .

f_{x} (x, y) = \frac{2 x}{x ^{2} + y ^{2}} + \frac{1}{1 + \frac{y ^{2}}{x ^{2}}} \cdot (- \frac{y}{x ^{2}}) = \frac{2 x}{x ^{2} + y ^{2}} - \frac{y}{x ^{2} + y ^{2}} = \frac{2 x - y}{x ^{2} + y ^{2}}

f_{y} (x, y) = \frac{2 y}{x ^{2} + y ^{2}} + \frac{1}{1 + \frac{y ^{2}}{x ^{2}}} \cdot \frac{1}{x} = \frac{2 y}{x ^{2} + y ^{2}} + \frac{x}{x ^{2} + y ^{2}} = \frac{2 y + x}{x ^{2} + y ^{2}}

\nabla f (1, 1) = (\frac{2 - 1}{2}, \frac{2 + 1}{2}) = (\frac{1}{2}, \frac{3}{2})

The logarithm is differentiated as a composite function: the derivative is the reciprocal of the argument times the derivative of the argument, so with respect to $x$ one obtains $\frac{2 x}{x ^{2} + y ^{2}}$ and with respect to $y$ one obtains $\frac{2 y}{x ^{2} + y ^{2}}$ . The arctangent is also composite: its derivative is $\frac{1}{1 + u ^{2}}$ times the derivative of $u = \frac{y}{x}$ , which is $- \frac{y}{x ^{2}}$ with respect to $x$ and $\frac{1}{x}$ with respect to $y$ . Multiplying numerator and denominator by $x^{2}$ , the factor $\frac{1}{1 + \frac{y ^{2}}{x ^{2}}}$ becomes $\frac{x ^{2}}{x ^{2} + y ^{2}}$ , so the two terms of each derivative have the same denominator $x^{2} + y^{2}$ and can be added. At the point $(1, 1)$ the denominator equals $2$ , therefore the gradient is $(\frac{1}{2}, \frac{3}{2})$ .