Differentiation of Implicit Functions

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition of Derivative of Implicit Functions
• 2. Formula and Steps
• 3. Solved Examples
• 4. Conclusion

Derivative of Implicit Functions:
An implicit function is a function where the dependent variable and the independent variable are mixed together. To differentiate an implicit function, we use implicit differentiation.

For an equation involving $x$ and $y$, if we have:

$F(x, y) = 0$,

then the derivative of $y$ with respect to $x$, denoted as $\dfrac{dy}{dx}$, is given by:

$\boxed{\frac{dy}{dx} = -\frac{\frac{d}{dx} F(x, y)}{\frac{d}{dx} F'(y, x)}}$

This means we differentiate both sides of the equation with respect to $x$, applying the chain rule to terms involving $y$ as a function of $x$.

Example 1: Differentiating an Implicit Function

Given $x^2 + y^2 = 25$, find $\dfrac{dy}{dx}$.

Solution:

Differentiate both sides with respect to $x$:

$\dfrac{d}{dx} (x^2) + \dfrac{d}{dx} (y^2) = \dfrac{d}{dx} (25)$

$2x + 2y \cdot \dfrac{dy}{dx} = 0$

Now, solve for $\frac{dy}{dx}$:

$2y \cdot \dfrac{dy}{dx} = -2x$

$\dfrac{dy}{dx} = \dfrac{-2x}{2y}$

Simplifying: $\boxed{\dfrac{dy}{dx} = \dfrac{-x}{y}}$

Example 2: Differentiating a More Complex Implicit Function

Given $x^3 + y^3 = 6xy$, find $\dfrac{dy}{dx}$.

Solution:

Differentiate both sides with respect to $x$:

$\dfrac{d}{dx} (x^3) + \dfrac{d}{dx} (y^3) = \dfrac{d}{dx} (6xy)$

Applying the chain rule:

$3x^2 + 3y^2 \cdot \dfrac{dy}{dx} = 6 \left( x \dfrac{d}{dx}(y) + y \right)$

Now, solving for $\frac{dy}{dx}$:

$3x^2 + 3y^2 \cdot \dfrac{dy}{dx} = 6x \cdot \dfrac{dy}{dx} + 6y$

Rearranging:

$3y^2 \cdot \dfrac{dy}{dx} - 6x \cdot \dfrac{dy}{dx} = 6y - 3x^2$

Factoring:

$\dfrac{dy}{dx} (3y^2 - 6x) = 6y - 3x^2$

Solving for $\dfrac{dy}{dx}$:

$\dfrac{dy}{dx} = \dfrac{6y - 3x^2}{3y^2 - 6x}$

Thus, the final answer is: $\boxed{\dfrac{dy}{dx} = \dfrac{6y - 3x^2}{3y^2 - 6x}}$.

Example 3: Differentiating a Trigonometric Implicit Function

Given $\sin(xy) = x + y$, find $\dfrac{dy}{dx}$.

Solution:

Differentiate both sides with respect to $x$:

$\dfrac{d}{dx} (\sin(xy)) = \dfrac{d}{dx} (x + y)$

Using the chain rule on the left side:

$\cos(xy) \cdot \left( y + x \cdot \dfrac{dy}{dx} \right) = 1 + \dfrac{dy}{dx}$

Now, isolate $\dfrac{dy}{dx}$:

$\cos(xy) \cdot y + x \cos(xy) \cdot \dfrac{dy}{dx} = 1 + \dfrac{dy}{dx}$

Rearranging:

$x \cos(xy) \cdot \dfrac{dy}{dx} - \dfrac{dy}{dx} = 1 - \cos(xy) \cdot y$

Factoring out $\frac{dy}{dx}$:

$\dfrac{dy}{dx} \cdot (x \cos(xy) - 1) = 1 - \cos(xy) \cdot y$

Solving for $\dfrac{dy}{dx}$:

$\dfrac{dy}{dx} = \dfrac{1 - \cos(xy) \cdot y}{x \cos(xy) - 1}$

Thus, the final answer is: $\boxed{\dfrac{dy}{dx} = \dfrac{1 - \cos(xy) \cdot y}{x \cos(xy) - 1}}$.

Example 4: Implicit Differentiation in a Real-World Context

The relationship between the height $h$ and radius $r$ of a cone is given by $h^2 + r^2 = 100$. Find $\dfrac{dh}{dr}$.

Solution:

Differentiate both sides with respect to $r$:

$\dfrac{d}{dr} (h^2) + \dfrac{d}{dr} (r^2) = \dfrac{d}{dr} (100)$

Using implicit differentiation:

$2h \cdot \dfrac{dh}{dr} + 2r = 0$

Now, solving for $\dfrac{dh}{dr}$:

$2h \cdot \dfrac{dh}{dr} = -2r$

$\dfrac{dh}{dr} = \dfrac{-r}{h}$

Thus, the final answer is: $\boxed{\dfrac{dh}{dr} = \dfrac{-r}{h}}$.