Derivatives of Implicit Functions

This lesson covers how to differentiate implicit functions, equations where $x$ and $y$ appear mixed together, and find $\tfrac{dy}{dx}$. It works through the general method and three examples of increasing difficulty.

What is an implicit function

An implicit function is one whose equation contains terms in both $x$ and $y$, rather than $y$ written on its own. For example:

$x^2 + xy + y^2 = 100$

$x + 2y = y^2 + 3x^3$

The method

To find $\tfrac{dy}{dx}$ for an implicit function:

• Differentiate each term with respect to $x$, and whenever you differentiate a term in $y$, multiply by $\tfrac{dy}{dx}$.
• Collect all the $\tfrac{dy}{dx}$ terms on the left and transpose the remaining terms to the right.
• Solve for $\tfrac{dy}{dx}$.

Example 1: a polynomial in x and y

Find $\tfrac{dy}{dx}$ for $x^2 + xy + y^2 = 100$.

Differentiate each term with respect to $x$, using the product rule on $xy$:

$2x + \left(y + x\tfrac{dy}{dx}\right) + 2y\tfrac{dy}{dx} = 0$

Collect the $\tfrac{dy}{dx}$ terms on the left:

$\tfrac{dy}{dx}(x + 2y) = -(2x + y)$

So:

$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$

Example 2: fractional powers

Given $x^{2/3} + y^{2/3} = a^{2/3}$, find $\tfrac{dy}{dx}$. Here $a$ is a constant.

Differentiate with respect to $x$:

$\tfrac{2}{3}x^{-1/3} + \tfrac{2}{3}y^{-1/3}\tfrac{dy}{dx} = 0$

Divide through by $\tfrac{2}{3}$ and keep the $\tfrac{dy}{dx}$ term on the left:

$y^{-1/3}\tfrac{dy}{dx} = -x^{-1/3}$

So:

$\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\frac{y^{1/3}}{x^{1/3}} = -\sqrt[3]{\frac{y}{x}}$

Example 3: square roots, simplify first

Given $x\sqrt{1+y} + y\sqrt{1+x} = 0$ with $x \neq y$, prove that $\tfrac{dy}{dx} = -\tfrac{1}{(1+x)^2}$.

The key idea is to simplify and find $y$ in terms of $x$ before differentiating. Move one term to the right:

$x\sqrt{1+y} = -y\sqrt{1+x}$

Square both sides to remove the roots:

$x^2(1+y) = y^2(1+x)$

Expand and bring everything to one side:

$x^2 - y^2 + x^2 y - x y^2 = 0$

Factorise. Write $x^2 - y^2 = (x-y)(x+y)$ and take $xy$ out of the remaining pair:

$(x-y)(x+y) + xy(x-y) = 0$

$(x-y)(x + y + xy) = 0$

Since $x \neq y$, the factor $x - y \neq 0$, so:

$x + y + xy = 0$

Solve for $y$ by taking $y$ common from the last two terms:

$y(1 + x) = -x \quad\Rightarrow\quad y = -\frac{x}{1+x}$

Now differentiate using the quotient rule (denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared):

$\frac{dy}{dx} = -\frac{(1+x)(1) - x(1)}{(1+x)^2} = -\frac{1}{(1+x)^2}$

as required.

Key takeaways

• Differentiate every term with respect to $x$ and attach $\tfrac{dy}{dx}$ to each differentiated $y$ term, then solve for $\tfrac{dy}{dx}$.
• Use the product rule on terms like $xy$ and the chain rule on powers of $y$.
• Sometimes it is easier to simplify the equation first, for instance squaring to clear roots, so you can write $y$ explicitly before differentiating.