Implicit Differentiation examples with detailed solutions are presented

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 \( \dfrac{dy}{dx} \).

- 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.

\( \dfrac{dy}{dx} = -\dfrac{y}{x + \cos y} \)

- 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 \( \dfrac{dy}{dx} \).

\( \dfrac{dy}{dx} = \dfrac{(-1 - y^{2})}{(4y^{3} + 2xy)} \)

where the tangent lines are parallel to the line \( x + y = 2 \)

- 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 \( (\sqrt{2} , \sqrt{2}) \)

1) \( x e^{y} = 3 \)

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

3) \( x \sin(x y) = x \)

2) \( \dfrac{dy}{dx} = - \dfrac{x}{y} \)

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

Rules of Differentiation of Functions in Calculus

differentiation and derivatives