Collecting dy/dx Terms Without Sign Errors

Why Collecting Terms Carefully Matters

When differentiating an implicit equation, \(y\) depends on \(x\), so the chain rule produces terms containing \(\frac{dy}{dx}\). Sometimes those terms appear in several places, including on both sides of the equation. The derivative is not isolated until all of its terms have been collected and their shared factor has been made explicit.

This step is algebra, but it is algebra with an important consequence: an incorrect sign or an unjustified division can change the derivative or make an expression appear to exist where it does not. A reliable method is to move all derivative terms to one side, move all remaining terms to the other, factor \(\frac{dy}{dx}\), and then check whether its coefficient is nonzero before dividing.

A General Form for Collecting Derivative Terms

Write \(y'=\frac{dy}{dx}\) as a compact alternative for the derivative. Suppose differentiation has produced an equation of the form

Here \(A,B,C,\) and \(D\) stand for expressions that do not contain \(y'\). Subtract \(C(x,y)y'\) from both sides and subtract \(B(x,y)\) from both sides. The result is

If \(A(x,y)-C(x,y)\ne 0\), divide both sides by that coefficient to obtain

The subtraction order matters. The coefficient of \(y'\) is the left-side coefficient minus the right-side coefficient. The terms without \(y'\) are moved in the opposite direction, so their new right-side expression is \(D-B\).

Why the Method Works

The reasoning is the ordinary algebraic rule for solving a linear equation in one unknown. At a particular point on an implicitly defined curve, regard \(x\) and \(y\) as fixed values and regard \(y'\) as the quantity to solve for. An equation \(K y'=R\) has the unique solution \(y'=R/K\) if \(K\ne 0\). Multiplying that proposed solution by \(K\) gives \(K(R/K)=R\), so it satisfies the equation. Uniqueness follows because if both \(u\) and \(v\) satisfy \(Ku=R\) and \(Kv=R\), then \(Ku=Kv\), and \(K\ne0\) implies \(u=v\).

In the general form above, \(K=A-C\) and \(R=D-B\). This explains both the rearrangement and the condition on division: if \(A-C=0\), division by that expression is not allowed. The collected equation then becomes \(0\cdot y'=D-B\). If \(D-B\ne0\), the equation is inconsistent at that point; if \(D-B=0\), it does not determine a unique value of \(y'\). Either way, the usual quotient formula cannot be used there.

A Reliable Collection Process

Use the following sequence whenever \(\frac{dy}{dx}\) occurs more than once. Keep each sign attached to its term while moving terms; do not change signs merely because a term looks complicated.

Worked Examples

In each example, the displayed equation is the result after differentiating an implicit equation. The task here is specifically to collect and isolate its derivative terms.

Common Sign and Division Errors

A frequent mistake is to move a term across the equals sign without changing its sign. For example, starting from \(5-2y'=3x y'-4\), one might incorrectly write \(5+4=(3x-2)y'\). This changes \(-2y'\) to \(-2y'\) on the right without correctly applying an operation to both sides. The correct collection is

The sign of the left-side term changes when it is moved to the right: \(-2y'\) becomes \(+2y'\). The constant \(-4\) moves to the left as \(+4\). Both changes follow from adding the appropriate terms to both sides.

Another error is to divide before collecting every derivative term. In \(4x y'-6=2y'+x\), dividing the left side by \(4x\) does not isolate \(y'\), because \(2y'\) remains on the right. First form \((4x-2)y'=x+6\), then divide only when \(4x-2\ne0\).

Finally, do not cancel terms that are added or subtracted. For instance, \((4x-2)y'=x+6\) does not simplify by cancelling the \(x\) in the coefficient with the \(x\) in the numerator. Cancellation applies to common factors in a quotient, not to terms joined by addition. The quotient \(\frac{x+6}{4x-2}\) should remain as written unless a valid factorization gives a common factor.

What Happens When the Coefficient Is Zero?

The condition on division is part of the answer, not a technical detail to ignore. In the third example, the collected equation is \((4x-2)y'=x+6\). At \(x=\frac12\), the coefficient \(4x-2\) is zero, while the other side is \(\frac12+6=\frac{13}{2}\). The equation would require \(0\cdot y'=\frac{13}{2}\), which has no solution for \(y'\). Therefore the quotient formula cannot be applied at that value.

More generally, after collecting terms into \(K y'=R\), first check \(K\). If \(K\ne0\), division gives a unique value. If \(K=0\) and \(R\ne0\), the equation cannot hold at that point. If both \(K=0\) and \(R=0\), the equation reduces to \(0=0\) and does not determine \(y'\). These cases explain why a denominator restriction must be stated whenever the derivative is written as a quotient.