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 dy/dx . As an example we know how to find 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 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 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 .
Implicit Differentiation Examples
Examples
Example 1
Use implicit differentiation to find the derivative dy / dx where y x + sin y = 1Solution 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[1]/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 = 3Solution 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[3] / 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 equationwhere 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 / x2) dy/dx =  x / y
3) dy/dx = [ 1  sin(x y) x y cos(x y) ] / [ x ^{2} cos(x y)]
More References and links
Tables of Formulas for DerivativesRules of Differentiation of Functions in Calculus
differentiation and derivatives