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.

Implicit Differentiation Examples

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[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 = 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[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 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) ] / [ x2 cos(x y)]

More References and links

Tables of Formulas for Derivatives
Rules of Differentiation of Functions in Calculus
differentiation and derivatives