The Implicit Differentiation Formulas Examples 1

Recall from The Implicit Differentiation Formulas page that if a single variable function $y = f(x)$ can be written implicitly as $F(x, y) = 0$ for all $x \in D(f)$ and $F$ is differentiable, then the derivative $\frac{dy}{dx}$ can be computed with the following formula:

(1)

\begin{align} \frac{dy}{dx} = - \frac{F_x (x, y)}{F_y (x, y)} \end{align}

Similarly, if a two variable function $z = f(x, y)$ can be written implicitly as $F(x, y, z) = 0$ for all $(x, y) \in D(f)$ and $F$ is differentiable, then the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ can be computed with the following formulas:

(2)

\begin{align} \quad \frac{\partial z}{\partial x} = - \frac{F_x (x, y, z)}{F_z (x, y, z)} \quad \quad \frac{\partial z}{\partial y} = - \frac{F_y (x, y, z)}{F_z (x, y, z)} \end{align}

We will now look at some examples of applying the formulas above.

Example 1

Let $x^2e^yy^2 + 2x \cos y = 0$. Find $\frac{dy}{dx}$.

Let $F(x, y) = x^2e^yy^2 + 2x \cos y$. To apply the first formula, we must compute the partial derivatives of $F$. We note that $F_x (x, y) = 2xe^yy^2 + 2 \cos y$ and $F_y (x, y) = x^2(e^yy^2 + 2e^yy) - 2x \sin y$. Therefore:

(3)

\begin{align} \quad \frac{dy}{dx} = - \frac{F_x (x, y)}{F_y (x, y)} = - \frac{2xe^yy^2 + 2 \cos y}{ x^2(e^yy^2 + 2e^yy) - 2x \sin y} \end{align}

Example 2

Let $y^3 \ln (x^2 \cos ^2 x + 2y^4) = x^2y^2$. Find $\frac{dy}{dx}$.

Let $F(x, y) = y^3 \ln (x^2 \cos ^2 x + 2y^4) - x^2y^2$. To apply the first formula, we must compute the partial derivatives of $F$. We note that:

(4)

\begin{align} \quad F_x (x, y) = y^3 \left ( \frac{2x \cos ^2 x - 2x^2 \cos x \sin x}{x^2 \cos ^2 x + 2y^4} \right ) - 2xy^2 \end{align}

(5)

\begin{align} \quad F_y (x, y) = 3y^2 \ln (x^2 \cos ^2 x + 2y^4) + y^3 \left ( \frac{8y^3}{x^2 \cos ^2 x + 2y^4} \right ) - 2x^2y \end{align}

Therefore:

(6)

\begin{align} \quad \frac{dy}{dx} = - \frac{F_x (x, y)}{F_y (x, y)} = - \frac{y^3 \left ( \frac{2x \cos ^2 x - 2x^2 \cos x \sin x}{x^2 \cos ^2 x + 2y^4} \right ) - 2xy^2}{3y^2 \ln (x^2 \cos ^2 x + 2y^4) + y^3 \left ( \frac{8y^3}{x^2 \cos ^2 x + 2y^4} \right ) - 2x^2y} \end{align}

Example 3

Let $2 \sin (xz^3) = - e^{xy} + z^3$. Find $\frac{\partial z}{\partial x}$.

Let $F(x, y, z) = 2 \sin (xz^3) + e^{xy} - z^3$. To apply the second formula, we need to calculate the partial derivatives $F_x (x, y, z)$ and $F_z (x, y, z)$ of $F$. We have that:

(7)

\begin{align} \quad F_x (x, y, z) = 2z^3 \cos (xz^3) + ye^{xy} \end{align}

(8)

\begin{align} \quad F_z (x, y, z) = 6xz^2 \cos (xz^3) - 3z^2 \end{align}

Therefore:

(9)

\begin{align} \quad \frac{\partial z}{\partial x} = -\frac{F_x (x, y, z)}{F_z (x, y, z)} = - \frac{2z^3 \cos (xz^3) + ye^{xy}}{6xz^2 \cos (xz^3) - 3z^2} \end{align}