# 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 $$\dfrac{dy}{dx}$$.

## 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.
1) $$x e^{y} = 3$$
2) $$x^{2} + y^{2} = 20$$
3) $$x \sin(x y) = x$$

### Solutions to the Above Exercises

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