Implicit Differentiation Step by Step (With Examples)

In implicit differentiation you differentiate every term with respect to $x$, but you remember that $y$ secretly depends on $x$. That means every $y$ term needs the chain rule.

$$\frac{d}{dx}\left[y^n\right] = ny^{n-1}\frac{dy}{dx}$$

Compare these side by side:

• $\frac{d}{dx}[x^3] = 3x^2$, because $x$ is the variable you are differentiating with respect to.
• $\frac{d}{dx}[y^3] = 3y^2 \frac{dy}{dx}$, because $y$ is an inside function of $x$.
• $\frac{d}{dx}[\sin y] = \cos y \, \frac{dy}{dx}$, for the same reason.
• $\frac{d}{dx}[25] = 0$, because constants still have a derivative of 0.

Think of it this way: whenever you take the derivative of something with $y$ in it, you tack on a $\frac{dy}{dx}$ (also written $y'$).

1. Differentiate both sides of the equation with respect to $x$.
2. Attach $\frac{dy}{dx}$ to every term that came from differentiating a $y$ expression. Use the product rule for terms like $xy$.
3. Collect every term containing $\frac{dy}{dx}$ on one side and everything else on the other side.
4. Factor out $\frac{dy}{dx}$ and divide to solve for it.

Your answer will usually contain both $x$ and $y$. That is normal. To get a number, plug in a point $(x, y)$ that lies on the curve.

Example: Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$, then find the tangent line at $(3, 4)$.

1. Differentiate both sides: $2x + 2y\frac{dy}{dx} = 0$.
2. Move the $x$ term over: $2y\frac{dy}{dx} = -2x$.
3. Divide by $2y$: $\frac{dy}{dx} = -\frac{x}{y}$.
4. Check that $(3, 4)$ is on the circle: $9 + 16 = 25$. It is.
5. Plug in the point: $\frac{dy}{dx} = -\frac{3}{4}$.
6. Use point-slope form for the tangent line: $y - 4 = -\frac{3}{4}(x - 3)$, which simplifies to $y = -\frac{3}{4}x + \frac{25}{4}$.

Sanity check: the radius from $(0, 0)$ to $(3, 4)$ has slope $\frac{4}{3}$, and a tangent to a circle is perpendicular to the radius. The negative reciprocal of $\frac{4}{3}$ is $-\frac{3}{4}$, which matches.

Terms like $6xy$ are a product of $x$ and $y$, so they need the product rule.

$$\frac{d}{dx}[xy] = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}$$

Example: Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$, then find the slope at $(3, 3)$.

1. Differentiate both sides: $3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$.
2. Collect the $\frac{dy}{dx}$ terms on the left: $3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$.
3. Factor: $\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$.
4. Divide and cancel the common factor of 3:

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

5. Check the point: $27 + 27 = 54$ and $6(3)(3) = 54$, so $(3, 3)$ is on the curve.
6. Plug in: $\frac{dy}{dx} = \frac{2(3) - 9}{9 - 2(3)} = \frac{-3}{3} = -1$.

Example: Find $\frac{dy}{dx}$ for $\sin(x + y) = y^2$.

1. The left side is a chain rule problem with inside $x + y$. The derivative of the inside is $1 + \frac{dy}{dx}$.
2. Differentiate both sides: $\cos(x + y)\left(1 + \frac{dy}{dx}\right) = 2y\frac{dy}{dx}$.
3. Distribute: $\cos(x + y) + \cos(x + y)\frac{dy}{dx} = 2y\frac{dy}{dx}$.
4. Collect the $\frac{dy}{dx}$ terms on the right: $\cos(x + y) = 2y\frac{dy}{dx} - \cos(x + y)\frac{dy}{dx}$.
5. Factor and divide:

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

To find $\frac{d^2y}{dx^2}$, differentiate your first derivative again, then substitute the first derivative back in.

Example: Find $\frac{d^2y}{dx^2}$ for $x^2 + y^2 = 25$.

1. From Example 1, $y' = -\frac{x}{y}$.
2. Use the quotient rule, remembering that $y$ still depends on $x$: $y'' = -\frac{(1)(y) - x \cdot y'}{y^2} = -\frac{y - xy'}{y^2}$.
3. Substitute $y' = -\frac{x}{y}$: $y'' = -\frac{y + \frac{x^2}{y}}{y^2}$.
4. Multiply the top and bottom by $y$: $y'' = -\frac{y^2 + x^2}{y^3}$.
5. Use the original equation, $x^2 + y^2 = 25$: $y'' = -\frac{25}{y^3}$.

That last substitution is a common trick. Whenever the original equation shows up inside your answer, replace it with its value.

Paste a curve into the VeilGPT solver to check your steps when the algebra gets long.