Solving Polynomial Relations For dy/dx

Why Polynomial Relations Need Care

A polynomial equation involving $x$ and $y$ does not always present $y$ explicitly as a formula in $x$. Even so, a differentiable function $y=y(x)$ may satisfy the equation as $x$ varies. Differentiating the relation gives an equation that can often be solved for the rate $dy/dx$.

The key point is that $y$ depends on $x$. Therefore, differentiating a term containing $y$ requires the chain rule. For instance, the derivative of $y^2$ is $2y\,dy/dx$, not merely $2y$. Likewise, the derivative of $xy$ includes a term containing $dy/dx$.

Key idea: Differentiate the relation with respect to $x$, treating $y$ as a differentiable function of $x$. Every differentiated occurrence of $y$ contributes a factor of $dy/dx$. Then collect those terms and isolate $dy/dx$, provided its coefficient is nonzero.

Differentiating Polynomial Terms

Suppose $y=y(x)$ is differentiable. The product rule gives

\frac{d}{dx}(xy) = \frac{d}{dx}(x)y+x\frac{dy}{dx} = y+x\frac{dy}{dx}.

The first term comes from differentiating $x$; the second comes from differentiating $y$. For a power of $y$, the chain rule gives

\frac{d}{dx}(y^n) = n y^{n-1}\frac{dy}{dx},

for a positive integer $n$. The factor $dy/dx$ appears because the input $y$ itself varies with $x$. Combining the product rule and chain rule gives, for nonnegative integers $a$ and $b$,

\frac{d}{dx}(x^a y^b) = a x^{a-1}y^b + b x^a y^{b-1}\frac{dy}{dx}.

If $a=0$, the first term is absent; if $b=0$, the second term is absent. In particular, a term with no $y$ contributes no factor of $dy/dx$. A term containing $y$ contributes such a factor when its $y$-dependence is differentiated.

Worked Example: Differentiate a Relation with a Product

Suppose $x^2+xy+y^2=1$, where $y$ is a differentiable function of $x$. Differentiate each term with respect to $x$:

\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(1).

The derivatives are $2x$, $y+x\,dy/dx$, $2y\,dy/dx$, and $0$, respectively. Thus

2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0.

Collect the terms containing $dy/dx$:

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

When $x+2y\ne 0$, division gives

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

For a check, the point $(1,0)$ satisfies the original relation because $1^2+1\cdot0+0^2=1$. The formula gives $dy/dx=-2/1=-2$ there. Substitution into the differentiated equation gives $2(1)+0+(1)(-2)+2(0)(-2)=0$, as required.

A General Collection Rule

Any polynomial relation can be written as $P(x,y)=0$, after moving all terms to one side. Here $P$ is a finite sum of terms of the form $c_{ab}x^a y^b$, where $c_{ab}$ is a constant and $a,b$ are nonnegative integers. If the original equation has a constant other than zero on its right-hand side, that constant is moved to the left before differentiating.

Theorem: Let $P(x,y)=\sum c_{ab}x^a y^b$ be a polynomial, and suppose a differentiable function $y=y(x)$ satisfies $P(x,y(x))=0$. Define $$ A(x,y)=\sum_{a\ge 1} a c_{ab}x^{a-1}y^b, \qquad B(x,y)=\sum_{b\ge 1} b c_{ab}x^a y^{b-1}, $$ where each sum runs over the terms of the polynomial with the indicated positive exponent. Then $$ A(x,y)+B(x,y)\frac{dy}{dx}=0. $$ At any point where $B(x,y)\ne 0$, the derivative is $$ \frac{dy}{dx}=-\frac{A(x,y)}{B(x,y)}. $$

Justification. Differentiate each term $c_{ab}x^a y^b$ using the product rule and chain rule. Its derivative is

c_{ab}a x^{a-1}y^b + c_{ab}b x^a y^{b-1}\frac{dy}{dx},

with a term omitted when its exponent multiplier is zero. Add these derivatives over the finite collection of polynomial terms. The terms not containing $dy/dx$ sum to $A(x,y)$; the coefficients of $dy/dx$ sum to $B(x,y)$. Since $P(x,y(x))$ is identically zero along the relation, its derivative is zero. Therefore $A+B\,dy/dx=0$. If $B\ne0$, ordinary algebra allows division by $B$, giving the stated formula.

This result explains both the method and its condition. Differentiation produces an equation linear in $dy/dx$, because each term of the first derivative contains at most one factor of $dy/dx$. The final division is valid only when the collected coefficient $B(x,y)$ is nonzero at the point under consideration.

Collecting All Contributions from $y$

When a term contains several factors of $y$, each occurrence matters. For example, $x^2y$ contributes $x^2\,dy/dx$ when its $y$-factor is differentiated. The term $xy^2$ contributes $2xy\,dy/dx$, because differentiating the square produces $2y\,dy/dx$, while differentiating its $x$-factor produces a separate term $y^2$.

Worked Example: A Cubic Relation

Consider $x^3+x^2y-y^3=1$, with $y$ a differentiable function of $x$. Differentiate each term:

3x^2+ \left(2xy+x^2\frac{dy}{dx}\right) - 3y^2\frac{dy}{dx} =0.

Here the product rule gives $d(x^2y)/dx=2xy+x^2\,dy/dx$, and the chain rule gives $d(y^3)/dx=3y^2\,dy/dx$. Collecting derivative terms and the remaining terms yields

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

Whenever $x^2-3y^2\ne0$,

\frac{dy}{dx} = -\frac{3x^2+2xy}{x^2-3y^2}.

At $(1,1)$, the original relation holds because $1+1-1=1$. The formula gives $dy/dx=-(3+2)/(1-3)=5/2$. Checking the differentiated equation at this point gives $3+2+(1-3)(5/2)=5-5=0$. Thus the computed value satisfies the differentiated relation.

Worked Example: Several Terms with $y$

Consider $x^2y+xy^2+y^3=3$. Differentiate term by term:

2xy+x^2\frac{dy}{dx} + y^2+2xy\frac{dy}{dx} + 3y^2\frac{dy}{dx} =0.

The first term, $x^2y$, contributes both $2xy$ and $x^2\,dy/dx$. The second, $xy^2$, contributes $y^2$ and $2xy\,dy/dx$. The last term contributes $3y^2\,dy/dx$. Collecting gives

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

At points where $x^2+2xy+3y^2\ne0$, this becomes

\frac{dy}{dx} = -\frac{2xy+y^2}{x^2+2xy+3y^2}.

The point $(1,1)$ satisfies the relation because $1+1+1=3$. The formula gives $dy/dx=-3/6=-1/2$. Substituting into the differentiated equation gives $2+1+(1+2+3)(-1/2)=3-3=0$, confirming the result.

A Common Mistake: Forgetting the Chain Rule

A frequent error is to differentiate a power of $y$ as though $y$ were a constant with respect to $x$. For the relation $x^2+xy+y^2=1$, an incorrect calculation might be

$$ 2x+y+2y=0. $$

This drops the derivative of $y$ in both $xy$ and $y^2$, so it cannot determine $dy/dx$. The correction is to apply the product rule to $xy$ and the chain rule to $y^2$:

2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0.

Only after every derivative term is present should they be collected and isolated. A useful check is to inspect each original term containing $y$: differentiating its dependence on $y$ must produce a factor of $dy/dx$.

When the Formula Can Be Used

The equation $A+B\,dy/dx=0$ follows along a differentiable solution of the polynomial relation. To solve it by division, the coefficient $B$ must be nonzero at the point in question. If $B=0$, division by $B$ is not permitted, and this algebraic step does not provide a value for $dy/dx$. The differentiated equation still has to be respected, but a separate analysis may be needed.

It is also important to keep track of what has been assumed: the relation is satisfied by a differentiable function $y(x)$. Differentiating a polynomial equation does not, by itself, establish that such a function exists at every point. For the calculations here, the relation and differentiability are the starting conditions.

Summary

To solve a polynomial relation for $dy/dx$, differentiate every term with respect to $x$. Use the product rule for products involving $x$ and $y$, and the chain rule whenever a power of $y$ is differentiated. Collect all terms containing $dy/dx$, move the remaining terms to the other side, and divide only if the coefficient of $dy/dx$ is nonzero.

Check Your Understanding

Use the product rule and chain rule explicitly in each question.

For $x^2+y^2=5$, differentiate the relation and isolate $dy/dx$, stating the condition for division.
For $x^2y+y=4$, write the differentiated equation and collect all terms containing $dy/dx$.
For $xy^2+x^3=2$, identify the separate contributions from differentiating $xy^2$.
In the general equation $A+B\,dy/dx=0$, what condition permits solving by division for $dy/dx$?
Explain why differentiating $y^4$ with respect to $x$ produces a factor of $dy/dx$.

Solving Polynomial Relations For dy/dx

What You'll Learn

Why Polynomial Relations Need Care

Differentiating Polynomial Terms

Worked Example: Differentiate a Relation with a Product

A General Collection Rule

Collecting All Contributions from \(y\)

Worked Example: A Cubic Relation

Worked Example: Several Terms with \(y\)

A Common Mistake: Forgetting the Chain Rule

When the Formula Can Be Used

Summary

Check Your Understanding

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Vol. 2: Solutions Manual

Solving Polynomial Relations For dy/dx

What You'll Learn

Why Polynomial Relations Need Care

Differentiating Polynomial Terms

Worked Example: Differentiate a Relation with a Product

A General Collection Rule

Collecting All Contributions from \(y\)

Worked Example: A Cubic Relation

Worked Example: Several Terms with \(y\)

A Common Mistake: Forgetting the Chain Rule

When the Formula Can Be Used

Summary

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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