Implicit Differentiation

Explicit vs. implicit

A function given as $y = f(x)$ is explicit. When $x$ and $y$ are tangled together in one equation instead, differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (so every $y$ term picks up a $\frac{dy}{dx}$ factor via the chain rule), then solve for $\frac{dy}{dx}$.

The method, step by step

1. Differentiate every term on both sides with respect to $x$.
2. Each time you differentiate a term containing $y$, multiply by $\dfrac{dy}{dx}$ (the chain rule): $\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}$.
3. Use the product rule on mixed terms: $\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}$.
4. Move every term containing $\dfrac{dy}{dx}$ to one side and everything else to the other.
5. Factor out $\dfrac{dy}{dx}$ and divide.

Worked example

Find $\dfrac{dy}{dx}$ if $2x + 3y = \sin x$.

Solution: Differentiate both sides with respect to $x$:

Worked example: a product term

Find $\dfrac{dy}{dx}$ if $x^2 + xy + y^2 = 7$.

Solution: The middle term needs the product rule:

Collecting the $\dfrac{dy}{dx}$ terms:

Second derivatives

To find $\dfrac{d^2y}{dx^2}$, differentiate the expression for $\dfrac{dy}{dx}$ again, then substitute the first derivative back in.

For the circle $x^2 + y^2 = 25$, find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$, and evaluate both at $(3, 4)$.

Solution: Differentiating once: $2x + 2y\dfrac{dy}{dx} = 0$, so $\dfrac{dy}{dx} = -\dfrac{x}{y}$. Differentiating $-\dfrac{x}{y}$ with the quotient rule (remembering $y$ depends on $x$):

The last step uses the original equation, $x^2 + y^2 = 25$. At $(3, 4)$: $\dfrac{dy}{dx} = -\dfrac{3}{4}$ and $\dfrac{d^2y}{dx^2} = -\dfrac{25}{64}$.

Worked example: an inverse trig identity

If $y = \cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)$, find $\dfrac{dy}{dx}$.

Solution: Substituting $x = \tan\theta$ turns the fraction into the double-angle identity $\dfrac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$, so:

Differentiating this simpler explicit form:

Common mistakes

• Leaving off $\dfrac{dy}{dx}$. $\dfrac{d}{dx}(y^3) = 3y^2\dfrac{dy}{dx}$, not $3y^2$. Every $y$ term picks up the factor.
• Treating $xy$ as a single variable. It is a product, so it needs the product rule: $y + x\dfrac{dy}{dx}$.
• Forgetting that constants differentiate to 0. In $x^2 + y^2 = 25$, the right side becomes 0, not 25.
• Not substituting back for a second derivative. $\dfrac{d^2y}{dx^2}$ usually still contains $\dfrac{dy}{dx}$ until you replace it with the first-derivative result.