Differentiation of Implicit Functions

1. What is an Implicit Function?

A function where y is not explicitly isolated in the form [y = f(x)], but instead appears mixed with x in an equation like:

• [x^{2} + y^{2} = 25]
• [xy + \sin y = 1]

Here, y is treated as a function of x, even if not directly solved for.

2. How do we differentiate implicitly?

Whenever we differentiate an expression containing y, we apply the Chain Rule, because:

y depends on x → its derivative must be multiplied by [\dfrac{dy}{dx}]

Examples:

• [\dfrac{d}{dx}(y) = \dfrac{dy}{dx}]
• [\dfrac{d}{dx}(y^{2}) = 2y\dfrac{dy}{dx}]
• [\dfrac{d}{dx}(\sin y) = \cos y \cdot \dfrac{dy}{dx}]
• [\dfrac{d}{dx}(e^{y}) = e^{y} \dfrac{dy}{dx}]

This is the key rule in implicit differentiation.

3. Examples (Step-by-Step Solutions)

Example 1

Differentiate [x^{2} + y^{2} = 25].

Step-by-Step Solution:

1. Differentiate both sides w.r.t [x]:
   [\dfrac{d}{dx}(x^{2}) + \dfrac{d}{dx}(y^{2})][ = \dfrac{d}{dx}(25)]
2. Apply derivatives:
   [2x + 2y\dfrac{dy}{dx} = 0]
3. Solve for [\dfrac{dy}{dx}]:
   [2y\dfrac{dy}{dx} = -2x]
   [\dfrac{dy}{dx} = -\dfrac{x}{y}]

Conclusion:
[\dfrac{dy}{dx} = -\dfrac{x}{y}]

Example 2

Differentiate [x\cdot y + \sin y = 2].

Step-by-Step Solution:

1. Differentiate term-by-term:
   [\dfrac{d}{dx}(xy) + \dfrac{d}{dx}(\sin y)][ = 0]
2. Use product rule for [xy] and chain rule for [\sin y]:
   [y + x\dfrac{dy}{dx} + \cos y \cdot \dfrac{dy}{dx}][ = 0]
3. Factor [\dfrac{dy}{dx}]:
   [y + \dfrac{dy}{dx}(x + \cos y)][ = 0]
4. Solve for [\dfrac{dy}{dx}]:
   [\dfrac{dy}{dx} = -\dfrac{y}{x + \cos y}]

Conclusion:
[\dfrac{dy}{dx}][ = -\dfrac{y}{x + \cos y}]

Example 3

Differentiate [e^{x+y} = x^{2} – y^{2}].

Step-by-Step Solution:

1. Differentiate both sides:
   [\dfrac{d}{dx}(e^{x+y})][ = \dfrac{d}{dx}(x^{2} – y^{2})]
2. Apply chain rule and power rule:
   [e^{x+y}(1 + \dfrac{dy}{dx})][ = 2x – 2y\dfrac{dy}{dx}]
3. Expand and rearrange:
   [e^{x+y} + e^{x+y}\dfrac{dy}{dx}][ = 2x – 2y\dfrac{dy}{dx}]
4. Group [\dfrac{dy}{dx}]:
   [e^{x+y}\dfrac{dy}{dx} + 2y\dfrac{dy}{dx}][ = 2x – e^{x+y}]
5. Factor out [\dfrac{dy}{dx}]:
   [\dfrac{dy}{dx}(e^{x+y} + 2y)][ = 2x – e^{x+y}]
6. Solve:
   [\dfrac{dy}{dx}][ = \dfrac{2x – e^{x+y}}{e^{x+y} + 2y}]

Conclusion:
[\dfrac{dy}{dx} = \dfrac{2x – e^{x+y}}{e^{x+y} + 2y}]

Example 4

Differentiate [\ln(x + y) = xy].

Step-by-Step Solution:

1. Differentiate both sides:
   [\dfrac{1}{x+y}(1 + \dfrac{dy}{dx})][ = y + x\dfrac{dy}{dx}]
2. Multiply both sides by [x+y]:
   [1 + \dfrac{dy}{dx}][ = (y + x\dfrac{dy}{dx})(x+y)]
3. Expand RHS:
   [1 + \dfrac{dy}{dx}][ = y(x+y) + x(x+y)\dfrac{dy}{dx}]
4. Collect [\dfrac{dy}{dx}] terms on one side:
   [\dfrac{dy}{dx} – x(x+y)\dfrac{dy}{dx}][ = y(x+y) – 1]
5. Factor out [\dfrac{dy}{dx}]:
   [\dfrac{dy}{dx}\big(1 – x(x+y)\big)][ = y(x+y) – 1]
6. Solve:
   [\dfrac{dy}{dx}][ = \dfrac{y(x+y) – 1}{1 – x(x+y)}]

Conclusion:
[\dfrac{dy}{dx}][ = \dfrac{y(x+y) – 1}{1 – x(x+y)}]