Implicit Differentiation

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 differentions to find the derivative \( \dfrac{dy}{dx} \). As an example we know how to find \( \dfrac{dy}{dx} \) if \( y = 2 x^3 - 2 x + 1 \).
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: \( x y + y^2 = 1 \) and asked to find \( \dfrac{dy}{dx} \).
The main idea of implicit differentiation is to differentiate both sides of the given equation 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 \( y x + \sin y = 1 \)
Solution to Example 1:
Differentiate both sides of the given equation and use the sum rule of differentiation to the whole term on the left of the given equation. \[ \dfrac{d}{dx}[xy] + \dfrac{d}{dx}[\sin y] = \dfrac{d}{dx}[1] \] Differentiate each term above using product rule to \( \dfrac{d}{dx}[x y] \) and the chain rule to \( \dfrac{d}{dx}[\sin y] \). \[ x \dfrac{dy}{dx} + y + \dfrac{dy}{dx} \cos(y) = 0 \] Note that in calculating \( \dfrac{d}{dx}[\sin y] \), we used the chain rule since \( y \) is itself a function of \( x \) and \( \sin (y) \) is a function of a function.
Solve for \( \dfrac{dy}{dx} \) to obtain. \[ \dfrac{dy}{dx} = -\dfrac{y}{x + \cos y} \]

Example 2

Use implicit differentiation to find the derivative \( \dfrac{dy}{dx} \) where \( y^{4} + x y^{2} + x = 3 \)
Solution to Example 2:
Use the differentiation of a sum formula to 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 above using 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 slope are equal to \( -1 \). The slope of tangent lines at a point can be found by implicity 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 \[ 2a + 2b (-1) = 0 \] Point \( P(a , b) \) is on the graph of \( x^{2} + y^{2} = 4 \), hence \[ a^{2} + b^{2} = 4 \] Solve the system of equations: \( 2a - 2b = 0 \) and \( a^{2} + b^{2} = 4 \) to obtain two points \[ (- \sqrt{2} , - \sqrt{2}) \) and \( (\sqrt{2} , \sqrt{2}) \]

Exercises

Use implicit differentiation to find \( \dfrac{dy}{dx} \) for each equation given below.

\( x e^{y} = 3 \)
\( x^{2} + y^{2} = 20 \)
\( x \sin(x y) = x \)

Solutions to the Above Exercises

\( \dfrac{dy}{dx} = - \dfrac{1}{x} \)
\( \dfrac{dy}{dx} = - \dfrac{x}{y} \)
\( \dfrac{dy}{dx} = \dfrac{1 - \sin(x y) -x y \cos(x y)}{x^{2} \cos(x y)} \)

Implicit Differentiation