Section 14.5 Chain Rule

We suppose that an equation of the form \(F(x, y) = 0\) defines \(y\) implicitly as a differentiable function of \(x\), that is, \(y = f(x)\), where \(F(x, f(x)) = 0\) for all \(x\) in the domain of \(f\). If \(F\) is differentiable, we can apply Case 1 of the Chain Rule to differentiate both sides of the equation \(F(x, y) = 0\) with respect to \(x\). Since both \(x\) and \(y\) are functions of \(x\), we obtain \[ \frac{\partial F}{\partial x} \frac{dx}{dx} + \frac{\partial F}{\partial y} \frac{dy}{dx} = 0 \] But \(dx/dx = 1\), so if \(\partial F / \partial y \ne 0\) we solve for \(dy/dx\) and obtain \[ \frac{dy}{dx} = - \frac{\partial F / \partial x}{\partial F / \partial y} = - \frac{F_x}{F_y} \]

Note

To derive this equation we assumed that \(F(x, y) = 0\) defines \(y\) implicitly as a function of \(x\). The Implicit Function Theorem, proved in advanced calculus, gives conditions under which this assumption is valid: it states that if \(F\) is defined on a disk containing \((a, b)\), where \(F(a, b) = 0, F_y(a, b) \ne 0\), and \(F_x\) and \(F_y\) are continuous on the disk, then the equation \(F(x, y) = 0\) defines \(y\) as a function of \(x\) near the point \((a, b)\) and the derivative of this function is given by Equation 6.

EXAMPLE 8 Find \(y'\) if \(x^3 + y^3 = 6xy\).

Now we suppose that \(z\) is given implicitly as a function \(z = f(x, y)\) by an equation of the form \(F(x, y, z) = 0\). This means that \(F(x, y, f(x, y)) = 0\) for all \((x, y)\) in the domain of \(f\). If \(F\) and \(f\) are differentiable, then we can use the Chain Rule to differentiate the equation \(F(x, y, z) = 0\) as follows: \[ \frac{\partial F}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial x} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0 \] But \[ \frac{\partial}{\partial x}(x) = 1 \quad \text{and} \quad \frac{\partial}{\partial x}(y) = 0 \] so this equation becomes \[ \frac{\partial F}{\partial x} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0 \] If \(\partial F / \partial z \ne 0\), we solve for \(\partial z / \partial x\) and obtain the first formula in Equations 7. The formula for \(\partial z / \partial y\) is obtained in a similar manner. \[ \frac{\partial z}{\partial x} = - \frac{\partial F / \partial x}{\partial F / \partial z} \quad \frac{\partial z}{\partial y} = - \frac{\partial F / \partial y}{\partial F / \partial z} \]

Again, a version of the Implicit Function Theorem stipulates conditions under which our assumption is valid: if \(F\) is defined within a sphere containing \((a, b, c)\), where \(F(a, b, c) = 0, F_z(a, b, c) \ne 0\), and \(F_x, F_y\), and \(F_z\) are continuous inside the sphere, then the equation \(F(x, y, z) = 0\) defines \(z\) as a function of \(x\) and \(y\) near the point \((a, b, c)\) and this function is differentiable, with partial derivatives given by (7).

EXAMPLE 9 Find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) if \(x^3 + y^3 + z^3 + 6xyz = 1\).