Implicit differentiation of an equation with x plus y cubed

Key concept: The equation 2x + (x+y)^3 = y becomes tractable once the chain rule is applied to the composite term (x+y)^3 [1].

Step by step

Differentiate each term with respect to x

For the equation $2x+(x+y)^3=y$, every term must be differentiated with respect to $x$, and $y$ is treated as a function of $x$. That means $\dfrac{d}{dx}(y)=\dfrac{dy}{dx}$, while $\dfrac{d}{dx}(2x)=2$.

The composite term $(x+y)^3$ requires the chain rule. Let the inside be $u=x+y$. Then

$\frac{d}{dx}(u^3)=3u^2\frac{du}{dx}=3(x+y)^2\left(1+\frac{dy}{dx}\right).$

So differentiating the whole equation gives

$2+3(x+y)^2\left(1+\frac{dy}{dx}\right)=\frac{dy}{dx}.$

Isolate the derivative

Expand the bracket to separate the $\dfrac{dy}{dx}$ terms:

$2+3(x+y)^2+3(x+y)^2\frac{dy}{dx}=\frac{dy}{dx}.$

Now move all derivative terms to one side:

$2+3(x+y)^2=\frac{dy}{dx}-3(x+y)^2\frac{dy}{dx}.$

Factor out $\dfrac{dy}{dx}$ on the right:

$2+3(x+y)^2=\frac{dy}{dx}\bigl(1-3(x+y)^2\bigr).$

Finally divide by $1-3(x+y)^2$:

$\frac{dy}{dx}=\frac{2+3(x+y)^2}{1-3(x+y)^2}.$

Final result

$\boxed{\frac{dy}{dx}=\frac{2+3(x+y)^2}{1-3(x+y)^2}}.$

This is already in terms of $x$ and $y$, which is exactly what implicit differentiation should produce.

Why the chain rule matters here

The term $(x+y)^3$ is not just a polynomial in $x$; it is a composite function whose inside also depends on $x$ through $y$. That is why the derivative of the inside is $1+\dfrac{dy}{dx}$, not just 1. Missing the $\dfrac{dy}{dx}$ part is one of the most common reasons students lose marks on implicit differentiation problems.

The algebra after differentiation is just as important as the calculus. Once the derivative appears on both sides, the goal is to collect all $\dfrac{dy}{dx}$ terms together and factor them cleanly. That step turns a long expression into a simple fraction.

Common mistake to avoid

Do not differentiate $(x+y)^3$ as $3(x+y)^2$ and stop there. That ignores the derivative of the inside function $x+y$, which contributes the extra factor $1+\dfrac{dy}{dx}$. Another frequent error is distributing incorrectly when expanding $3(x+y)^2(1+\dfrac{dy}{dx})$; both the constant part and the derivative part must be kept. A final trap is misplacing the derivative after factoring. When you move $3(x+y)^2\dfrac{dy}{dx}$ to the other side, the sign changes, so the denominator must become $1-3(x+y)^2$, not $1+3(x+y)^2$.

If you keep the chain rule, product structure, and rearrangement in separate stages, the result is straightforward to verify.

Pitfall alert

The hardest step in this implicit differentiation problem is the composite term $(x+y)^3$, because it is easy to differentiate only the outer power and forget that the inside $x+y$ also changes with $x$. That missing factor is not just 1; it is $1+dy/dx$. If you write $3(x+y)^2$ without the chain-rule multiplier, the final slope formula will be wrong even if the algebra afterward is perfect. Another place where students get stuck is the rearrangement stage. Once $dy/dx$ appears on both sides, you must gather every derivative term on one side before factoring. If you try to divide too early, the expression becomes harder to track and sign errors are likely. Also, because the answer is requested in terms of $x$ and $y$, do not try to solve for y explicitly unless the problem asks for it; implicit differentiation is meant to leave the relation intact.

Try different conditions

If the equation were changed to $2x+(x+y)^4=y$, the setup would be similar but the chain rule would produce a fourth-power derivative: $2+4(x+y)^3(1+dy/dx)=dy/dx$. Rearranging would give $2+4(x+y)^3+4(x+y)^3dy/dx=dy/dx$, so the derivative would become $dy/dx=\dfrac{2+4(x+y)^3}{1-4(x+y)^3}$. Another useful variation is $2x+(x-y)^3=y$. In that case the inside derivative changes to $1-dy/dx$ instead of $1+dy/dx$, because the derivative of $x-y$ is $1-dy/dx$. That sign change often alters the denominator after rearrangement, which is why the chain rule step must match the exact inside expression.

Further reading

implicit differentiation chain rule, solving for dy dx, composite function derivative