 # 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 derivativedy/dx. As an example we know how to find dy/dx if y = 2 x3 - 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 dy/dx. For example, we are given an equation that relates y and x as follows: x y + y2 = 1 and asked to find 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
dy/dx.

## Examples

### Example 1

Use implicit differentiation to find the derivative 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.
d [xy] / dx + d [siny] / dx = d/dx .
• Differentiate each term above using product rule to d [x y] / dx and the cain rule to d [sin y] / dx.
x dy / dx + y + (dy / dx) cos(y) = 0 .
• Note that in calculating d [siny] / dx, we used the chain rule since y is itself a function of x and sin (y) is a function of a function.
• Solve for dy/dx to obtain.
dy / dx = -y / (x + cos y)

### Example 2

Use implicit differentiation to find the derivative 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.
d[y 4] / dx + d[x y 2] / dx + d[x] / dx = d / dx
• Differentiate each term above using power rule, product rule and chain rule.
4y 3 dy / dx + (1) y 2 + x 2y dy / dx + 1 = 0
• Solve for dy/dx.
dy/dx = (-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 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 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
(- √2 , - √2) and (√2 , √2)

### Exercises

Use implicit differentiation to find 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) dy/dx = - 1 / x
2) dy/dx = - x / y
3) dy/dx = [ 1 - sin(x y) -x y cos(x y) ] / [ x
2 cos(x y)] 