Implicit differentiation examples with detailed solutions are presented.
Implicit Differentiation Explained
When we are given a function \( y \) explicitly in terms of \( x \), we use the rules and formulas of differentiation to find the derivative \( \dfrac{dy}{dx} \). As an example, we know how to find \( \dfrac{dy}{dx} \) if \( y = 2x^3 - 2x + 1 \).
In some other situations, however, instead of a function given explicitly, we are given an equation including terms in \( y \) and \( x \) and we are asked to find \( \dfrac{dy}{dx} \). For example, we are given an equation that relates \( y \) and \( x \) as follows: \( xy + y^2 = 1 \) and asked to find \( \dfrac{dy}{dx} \).
The main idea of implicit differentiation is to differentiate both sides of the given equation with respect to \( x \) and then solve the new equation obtained to find \( \dfrac{dy}{dx} \).
Implicit Differentiation Examples
Example 1
Use implicit differentiation to find the derivative \( \dfrac{dy}{dx} \) where \( yx + \sin y = 1 \)
Solution to Example 1
Differentiate both sides of the given equation and use the sum rule of differentiation:
\[ \dfrac{d}{dx}[xy] + \dfrac{d}{dx}[\sin y] = \dfrac{d}{dx}[1] \]Differentiate each term using the product rule for \( \dfrac{d}{dx}[xy] \) and the chain rule for \( \dfrac{d}{dx}[\sin y] \):
\[ x \dfrac{dy}{dx} + y + \dfrac{dy}{dx} \cos(y) = 0 \]Note that in calculating \( \dfrac{d}{dx}[\sin y] \), we used the chain rule since \( y \) is itself a function of \( x \).
Solve for \( \dfrac{dy}{dx} \):
\[ \dfrac{dy}{dx} = -\dfrac{y}{x + \cos y} \]Example 2
Use implicit differentiation to find the derivative \( \dfrac{dy}{dx} \) where \( y^{4} + xy^{2} + x = 3 \)
Solution to Example 2
Use the differentiation of a sum formula on the left side of the given equation:
\[ \dfrac{d[y^{4}]}{dx} + \dfrac{d[xy^{2}]}{dx} + \dfrac{d[x]}{dx} = \dfrac{d[3]}{dx} \]Differentiate each term using the power rule, product rule, and chain rule:
\[ 4y^{3} \dfrac{dy}{dx} + (1)y^{2} + x(2y \dfrac{dy}{dx}) + 1 = 0 \]Solve for \( \dfrac{dy}{dx} \):
\[ \dfrac{dy}{dx} = \dfrac{-1 - y^{2}}{4y^{3} + 2xy} \]Example 3
Find all points on the graph of the equation:
\[ x^{2} + y^{2} = 4 \]where the tangent lines are parallel to the line \( x + y = 2 \)
Solution to Example 3
Rewrite the given line \( x + y = 2 \) in slope-intercept form: \( y = -x + 2 \), and identify the slope as \( m = -1 \). The tangent lines are parallel to this line and therefore their slopes are equal to \( -1 \). The slope of tangent lines at a point can be found by implicit differentiation of \( x^{2} + y^{2} = 4 \):
\[ 2x + 2y \dfrac{dy}{dx} = 0 \]Let \( P(a, b) \) be the point of tangency. At point \( P \), the slope is \( -1 \). Substituting \( x \) by \( a \), \( y \) by \( b \), and \( \dfrac{dy}{dx} \) by \( -1 \) in the above equation, we obtain:
\[ 2a + 2b(-1) = 0 \implies 2a = 2b \implies a = b \]Point \( P(a, b) \) is on the graph of \( x^{2} + y^{2} = 4 \), hence:
\[ a^{2} + b^{2} = 4 \]Substitute \( a = b \) into \( a^2 + b^2 = 4 \):
\[ 2a^2 = 4 \implies a^2 = 2 \implies a = \pm\sqrt{2} \]Thus, we obtain two points:
\[ (-\sqrt{2}, -\sqrt{2}) \quad \text{and} \quad (\sqrt{2}, \sqrt{2}) \]Exercises
Use implicit differentiation to find \( \dfrac{dy}{dx} \) for each equation given below:
- \( xe^{y} = 3 \)
- \( x^{2} + y^{2} = 20 \)
- \( x \sin(xy) = x \)
Solutions to the Above Exercises
- \( \dfrac{dy}{dx} = -\dfrac{1}{x} \)
- \( \dfrac{dy}{dx} = -\dfrac{x}{y} \)
- \( \dfrac{dy}{dx} = \dfrac{1 - \sin(xy) - xy \cos(xy)}{x^{2} \cos(xy)} \)