Not every curve gives up $y$ in terms of $x$ — differentiate the equation as it stands, and let the chain rule carry the weight.
9 min read4 worked examples
Differentiating equations that aren't solved for y
Equations like $y=3x^2+2$ are explicit: $y$ is written directly in terms of $x$, ready to differentiate term by term. But an equation like $x^2+y^2=25$ is implicit — $y$ isn't isolated, and often can't be without splitting into multiple branches (here, $y=\pm\sqrt{25-x^2}$).
Implicit differentiation avoids rearranging altogether. Treat $y$ as an unspecified function of $x$, differentiate both sides of the equation term by term with respect to $x$, and apply the chain rule to any term involving $y$ — since $y$ itself depends on $x$.
Every time a term contains $y$, differentiating it with respect to $x$ requires the chain rule: differentiate with respect to $y$ first, then multiply by $\dfrac{dy}{dx}$. So $\dfrac{d}{dx}(y^n)=ny^{n-1}\dfrac{dy}{dx}$, and any term mixing $x$ and $y$ (like $xy$ or $x^2y^3$) additionally needs the product rule.
After differentiating every term, $\dfrac{dy}{dx}$ appears scattered throughout the equation. Collect all terms containing $\dfrac{dy}{dx}$ on one side, factor it out, and divide to make $\dfrac{dy}{dx}$ the subject — usually leaving an expression in terms of both $x$ and $y$.
Tangents and normals to implicit curves
Once $\dfrac{dy}{dx}$ has been found implicitly, finding a tangent or normal at a given point $(x_1,y_1)$ on the curve works exactly as before: substitute the point's coordinates into the (typically mixed) expression for $\dfrac{dy}{dx}$ to get a numerical gradient, then use $y-y_1=m(x-x_1)$ — with $m$ itself for the tangent, or $-\dfrac{1}{m}$ for the normal.
Worked example 1
Differentiating a circle implicitly
Find $\dfrac{dy}{dx}$ for the curve $x^2+y^2=25$.
MethodDifferentiate term by term, applying the chain rule to the y² term
1
Differentiate both sides with respect to $x$: $\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2)=\dfrac{d}{dx}(25)$.
Differentiate every term in the equation, left and right.
2
$2x+2y\dfrac{dy}{dx}=0$.
$x^2$ differentiates normally; $y^2$ needs the chain rule, giving $2y\frac{dy}{dx}$; the constant 25 differentiates to 0.
On the circle $x^2+y^2=25$, the gradient $-\dfrac{x}{y}$ is the negative reciprocal of $\dfrac{y}{x}$ — exactly what's expected, since the radius to any point has gradient $\dfrac{y}{x}$, and a tangent to a circle is always perpendicular to the radius at that point.
Worked example 2
A curve mixing x and y (needs the product rule too)
Find $\dfrac{dy}{dx}$ for the curve $x^2+3xy+y^2=9$.
MethodDifferentiate term by term; the xy term needs the product rule
At $(0,1)$: $\dfrac{dy}{dx}=\dfrac{e^0-1}{3(1)^2+0}=\dfrac{1-1}{3}=0$.
Substitute the given point.
5
The tangent gradient is $0$, so the normal is vertical: $x=0$.
A normal perpendicular to a horizontal tangent is a vertical line through the same point — the negative-reciprocal formula breaks down here since dividing by 0 is undefined, so this case must be reasoned about directly.
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Look deeper
Whenever an implicit tangent gradient comes out as $0$, the normal isn't found by "flipping and negating" $0$ (which is undefined) — it's simply the vertical line $x=x_1$ through the point. Always check for this special case before applying $-\dfrac{1}{m}$ automatically.
Common mistake
Forgetting to multiply by $\dfrac{dy}{dx}$ when differentiating a term containing $y$ — e.g. writing $\dfrac{d}{dx}(y^2)=2y$ instead of $2y\dfrac{dy}{dx}$. Also common: treating a mixed term like $xy$ as if only one variable needs differentiating, instead of applying the full product rule.
Check yourself
Find $\dfrac{dy}{dx}$ for the curve $x^3+y^3=6xy$. (Hint: the right-hand side is a product of x and y, and needs the product rule.)