Worked examples · Sheet 2

Studying differentiability with the definition

Theory used: Differentiable function Differentiability implies continuity Total differential theorem

Exercise 1

Verify with the definition that

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

is differentiable at $(0, 0)$ .

f_{x} (0, 0) = t \to 0 lim \frac{f ( t , 0 ) - f ( 0 , 0 )}{t} = t \to 0 lim \frac{t ^{2}}{t} = 0

f_{y} (0, 0) = t \to 0 lim \frac{f ( 0 , t ) - f ( 0 , 0 )}{t} = t \to 0 lim \frac{3 t ^{2}}{t} = 0

0 \leq \frac{f ( h , k ) - f ( 0 , 0 ) - 0 \cdot h - 0 \cdot k}{h ^{2} + k ^{2}} = \frac{h ^{2} + 3 k ^{2}}{h ^{2} + k ^{2}} \leq \frac{3 ( h ^{2} + k ^{2} )}{h ^{2} + k ^{2}} = 3 h^{2} + k^{2}

(h, k) \to (0, 0) lim 3 h^{2} + k^{2} = 0 ⟹ f is differentiable at (0, 0)

The first step is computing the partial derivatives at the origin with the definition: along the $x$ -axis the function equals $t^{2}$ and the difference quotient $\frac{t ^{2}}{t} = t$ tends to $0$ , so $f_{x} (0, 0) = 0$ ; along the $y$ -axis the quotient equals $3 t$ and tends to $0$ , so $f_{y} (0, 0) = 0$ . The second step is the limit in the definition of differentiability, with numerator $f (h, k) - f (0, 0) - f_{x} h - f_{y} k = h^{2} + 3 k^{2}$ . Since $h^{2} + 3 k^{2} \leq 3 (h^{2} + k^{2})$ , the ratio lies between $0$ and $3 h^{2} + k^{2}$ , which tends to $0$ ; by the comparison theorem the limit equals $0$ , so the definition is satisfied and $f$ is differentiable at the origin.

Exercise 2

Study the differentiability at $(0, 0)$ of

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

\frac{f ( t , 0 ) - f ( 0 , 0 )}{t} = \frac{t ^{2}}{t} = \frac{∣ t ∣}{t}

t \to 0^{+} lim \frac{∣ t ∣}{t} = 1, t \to 0^{-} lim \frac{∣ t ∣}{t} = - 1

f_{x} (0, 0) does not exist ⟹ f is not differentiable at (0, 0)

One starts from the difference quotient of the partial derivative with respect to $x$ : along the $x$ -axis the function equals $t^{2} = ∣ t ∣$ , so the quotient is $\frac{∣ t ∣}{t}$ . The one-sided limits equal $1$ and $- 1$ and are different, therefore the limit does not exist and $f$ is not partially differentiable with respect to $x$ at the origin. Since differentiability requires the existence of both partial derivatives, the necessary condition is not satisfied and $f$ is not differentiable at $(0, 0)$ . The graph of $f$ is a cone with vertex at the origin, a point at which no plane approximates the graph to first order.

Exercise 3

Study the differentiability at $(0, 0)$ of

f (x, y) = ⎩ ⎨ ⎧ \frac{x y}{x ^{2} + y ^{2}} 0 (x, y) \neq = (0, 0) (x, y) = (0, 0) .

f_{x} (0, 0) = t \to 0 lim \frac{f ( t , 0 ) - 0}{t} = t \to 0 lim \frac{0}{t} = 0, f_{y} (0, 0) = 0

\frac{f ( h , k ) - 0 - 0 \cdot h - 0 \cdot k}{h ^{2} + k ^{2}} = \frac{hk}{h ^{2} + k ^{2}}

along k = h : \frac{h ^{2}}{2 h ^{2}} = \frac{1}{2}, along k = 0 : 0

the limit does not exist ⟹ f is not differentiable at (0, 0)

On the axes the function is identically zero, because the numerator $x y$ vanishes when one of the two coordinates is zero; hence both difference quotients are zero and the partial derivatives at the origin exist and equal $0$ . The differentiability limit has the function $\frac{hk}{h ^{2} + k ^{2}}$ as its ratio. Along the line $k = h$ the ratio is constantly $\frac{1}{2}$ , while along the axis $k = 0$ it equals $0$ : two restrictions have different limits, so the limit does not exist and in particular it does not equal $0$ . The definition of differentiability is not satisfied, therefore $f$ is not differentiable at the origin, even though it is partially differentiable and also continuous, since $∣ f (x, y) ∣ \leq \frac{1}{2} x^{2} + y^{2}$ by the inequality $2∣ x y ∣ \leq x^{2} + y^{2}$ .