Worked examples · Sheet 1

First derivatives of implicit functions

Theory used: Dini's theorem (two variables)Implicitly defined function

Exercise 1

Verify that the equation

x^{3} + y^{3} - 2 x y = 0

implicitly defines a function $y = φ (x)$ near the point $(1, 1)$ and compute $φ^{'} (1)$ .

F (x, y) = x^{3} + y^{3} - 2 x y, F (1, 1) = 1 + 1 - 2 = 0

F_{y} (x, y) = 3 y^{2} - 2 x, F_{y} (1, 1) = 3 - 2 = 1 \neq = 0

F_{x} (x, y) = 3 x^{2} - 2 y, F_{x} (1, 1) = 3 - 2 = 1

φ^{'} (1) = - \frac{F _{x} ( 1 , 1 )}{F _{y} ( 1 , 1 )} = - \frac{1}{1} = - 1

The curve

x^{3} + y^{3} - 2 x y = 0

and the tangent line

y = 2 - x

at the point

(1, 1)

(dashed), with slope

φ^{'} (1) = - 1

We check the two hypotheses of Dini’s theorem at the point. The point belongs to the curve, since $F (1, 1) = 0$ ; the partial derivative with respect to $y$ at the point is $F_{y} (1, 1) = 1$ , which is nonzero, therefore the theorem applies and near $(1, 1)$ the curve is the graph of a function $y = φ (x)$ of class $C^{1}$ with $φ (1) = 1$ . The derivative is computed with the formula of the theorem, which requires only the partial derivatives of $F$ at the point: since $F_{x} (1, 1) = 1$ and $F_{y} (1, 1) = 1$ , we get $φ^{'} (1) = - 1$ .

Exercise 2

Verify that the equation

e^{x y} + x^{2} - y - 2 = 0

implicitly defines a function $y = φ (x)$ near the point $(0, - 1)$ and compute $φ^{'} (0)$ .

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

F_{y} (x, y) = x e^{x y} - 1, F_{y} (0, - 1) = 0 - 1 = - 1 \neq = 0

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

φ^{'} (0) = - \frac{F _{x} ( 0 , - 1 )}{F _{y} ( 0 , - 1 )} = - \frac{- 1}{- 1} = - 1

The point $(0, - 1)$ belongs to the curve because, after substitution, the exponential equals $e^{0} = 1$ and the sum $1 + 0 + 1 - 2$ is zero. To differentiate $e^{x y}$ with respect to $y$ we apply the chain rule: the derivative of the exponent $x y$ with respect to $y$ is $x$ , hence $\frac{\partial}{\partial y} e^{x y} = x e^{x y}$ ; at the point this term is zero and what remains is $F_{y} (0, - 1) = - 1$ , which is nonzero, therefore Dini’s theorem applies. In the same way $\frac{\partial}{\partial x} e^{x y} = y e^{x y}$ , which equals $- 1$ at the point. The derivative formula gives $φ^{'} (0) = - \frac{- 1}{- 1} = - 1$ .

Exercise 3

Verify that the equation

x ln y + y e^{x - 1} - 1 = 0

implicitly defines a function $y = φ (x)$ near the point $(1, 1)$ and compute $φ^{'} (1)$ .

F (x, y) = x ln y + y e^{x - 1} - 1, F (1, 1) = 1 \cdot 0 + 1 \cdot 1 - 1 = 0

F_{y} (x, y) = \frac{x}{y} + e^{x - 1}, F_{y} (1, 1) = 1 + 1 = 2 \neq = 0

F_{x} (x, y) = ln y + y e^{x - 1}, F_{x} (1, 1) = 0 + 1 = 1

φ^{'} (1) = - \frac{F _{x} ( 1 , 1 )}{F _{y} ( 1 , 1 )} = - \frac{1}{2}

The function $F$ is of class $C^{1}$ on the open set ${y > 0}$ , which contains the point $(1, 1)$ . At the point the logarithm vanishes and the exponential equals $e^{0} = 1$ , hence $F (1, 1) = 0$ and the point belongs to the curve. The derivative with respect to $y$ has two terms, $\frac{x}{y}$ from the term $x ln y$ and $e^{x - 1}$ from the term $y e^{x - 1}$ ; at the point their sum equals $2$ , which is nonzero, therefore Dini’s theorem applies. In the derivative with respect to $x$ the first term is $ln y$ , which is zero at the point, and the second is $y e^{x - 1}$ , which equals $1$ . From the formula of the theorem it follows that $φ^{'} (1) = - \frac{1}{2}$ .