Why Radicals Require Domain Checks
Implicit equations often combine \(x\) and \(y\) inside a radical, as in \(\sqrt{x+y}=x-y\). Differentiation still uses the chain rule, but a radical imposes a real-domain restriction, and its derivative formula has a more limited domain than the radical itself.
For example, \(\sqrt{u}\) is defined for real \(u\ge 0\), but its derivative formula \(1/(2\sqrt{u})\) is defined only for \(u>0\). The distinction matters: an equation may have points where a radicand is zero even though the usual implicit differentiation steps do not apply there.
Real Domains of Radicals and Fractional Powers
An even root of a real expression requires a nonnegative radicand. Thus \(\sqrt{g(x,y)}\) is real where \(g(x,y)\ge 0\), and \(\sqrt[4]{g(x,y)}\) has the same nonnegativity requirement. An odd root, such as \(\sqrt[3]{g(x,y)}\), is real for every real value of its radicand.
When an even root is differentiated by the ordinary chain rule, the radicand must be positive. If \(u=u(x)\) is differentiable and \(u(x)>0\), then
For a positive radicand, fractional powers can be differentiated in the same way as other powers. For example, \(\sqrt[3]{u}=u^{1/3}\) has derivative \(\frac{1}{3}u^{-2/3}u'\) where \(u\ne 0\). The formula is not defined at \(u=0\), so it cannot be used there without a separate analysis.
In an implicit equation, \(y\) is treated as a function of \(x\). Consequently, the chain rule must include the derivative of every expression involving \(y\). For instance, where \(x+y(x)>0\),
The numerator \(1+y'\) comes from differentiating the entire radicand \(x+y(x)\), not just its \(x\)-term.
Deriving the Square-Root Rule
The square-root derivative formula follows directly from a difference quotient. Let \(g\) be differentiable at \(x\), with \(g(x)>0\). For sufficiently small \(h\), continuity ensures that \(g(x+h)>0\) as well. Rationalizing gives
As \(h\) approaches zero, the numerator divided by \(h\) approaches \(g'(x)\), and the sum in the denominator approaches \(2\sqrt{g(x)}\). Because \(g(x)>0\), this denominator is nonzero. Therefore,
This argument also explains the strict positivity condition: if \(g(x)=0\), the limiting denominator in this calculation is zero, so the argument does not establish the formula there.
Worked Examples
Worked Example: Differentiating a Square Root of a Two-Variable Expression
Differentiate the relation \(\sqrt{x+y}=x-y\) implicitly, stating where the calculation applies.
First, the original radical requires \(x+y\ge 0\). To use the derivative formula, restrict the calculation to \(x+y>0\). Differentiate each side with respect to \(x\), treating \(y\) as a function of \(x\):
Multiply both sides by \(2\sqrt{x+y}\), which is positive on this domain:
Collect the terms containing \(y'\):
Factor and divide by the positive coefficient \(1+2\sqrt{x+y}\):
Check by substituting this expression into the differentiated equation. Writing \(s=\sqrt{x+y}\), the left side is \((1+y')/(2s)\). Since \(y'=(2s-1)/(1+2s)\), we have \(1+y'=4s/(1+2s)\), so the left side is \(2/(1+2s)\). The right side is \(1-y'=2/(1+2s)\). The two sides agree. This formula was derived only for \(x+y>0\), not at the boundary \(x+y=0\).
Worked Example: A Square Root Whose Radicand Is \(y\)
Differentiate \(x^2+\sqrt{y}=3\) implicitly wherever the usual square-root derivative applies, and check the result.
The equation requires \(y\ge 0\). For the direct differentiation, take \(y>0\). Differentiating gives
Multiply by \(2\sqrt{y}\):
Therefore, on the part of the relation where \(y>0\),
To check, the original equation gives \(\sqrt{y}=3-x^2\), so \(y=(3-x^2)^2\), with \(3-x^2\ge 0\) required by the square root. Differentiating this expression gives
On the original relation, \(3-x^2=\sqrt{y}\), so this is exactly \(y'=-4x\sqrt{y}\). At \(y=0\), the implicit differentiation step that divides by \(2\sqrt{y}\) is unavailable; the derivative formula above must not be extended to those points by substitution.
Worked Example: Squaring Can Add Points
Consider \(\sqrt{x-y}=y\). Differentiate the original equation, then compare it with the equation obtained by squaring.
The radical requires \(x-y\ge 0\). Also, since a square root is nonnegative, the original equation requires \(y\ge 0\). For differentiation by the usual rule, take \(x-y>0\). Differentiating the original equation gives
Multiply by \(2\sqrt{x-y}\), then collect the \(y'\) terms:
The coefficient is positive, so
As a check, squaring the original equation gives \(x-y=y^2\), or \(x=y^2+y\). Differentiating this equation gives \(1=(2y+1)y'\). On points that satisfy the original equation, \(\sqrt{x-y}=y\), so the derivative formula becomes \(y'=1/(1+2y)\), which satisfies \(1=(2y+1)y'\).
However, the squared equation alone is not equivalent to the original: it does not retain the condition \(y\ge 0\). For example, \((x,y)=(0,-1)\) satisfies the squared equation because \(x-y=1=y^2\), but it fails the original equation: \(\sqrt{0-(-1)}=1\ne -1\). Squaring has introduced an extraneous point.
Preserving the Original Relation
Squaring both sides is a common way to remove a radical, but it generally preserves only one direction of equivalence. If \(\sqrt{A}=B\), then squaring implies \(A=B^2\). The reverse implication requires additional conditions: \(A\ge 0\) and \(B\ge 0\). Without the sign condition on \(B\), the squared equation may describe extra points.
Likewise, raising both sides to an even power can discard sign information. Dividing by an expression can discard points where that expression is zero. These transformations can be useful, but the domain and any excluded cases must be recorded before using them.
Common Mistake: Using the Formula at a Boundary
A frequent error is to apply the square-root derivative formula at a point where the radicand is zero. For example, from \(\frac{d}{dx}\sqrt{g(x)}=\frac{g'(x)}{2\sqrt{g(x)}}\), substituting \(g(x)=0\) produces a zero denominator. That does not give a derivative; it shows that this formula cannot be used there.
The correction is to separate the domain check from the differentiation. First identify where the radical is defined, then identify where its derivative formula applies. At a boundary point, a derivative might still exist for a particular relation or branch, but it must be established by a separate argument, not by using a formula whose denominator is zero.
A Reliable Procedure
For each even root, require its radicand to be nonnegative. Include sign restrictions implied by the original equation.
For the usual even-root derivative formula, require the radicand to be strictly positive.
Use the chain rule on each radicand, including every \(y'\) term.
Before dividing, check that the coefficient or factor is nonzero. If you square or otherwise transform the equation, preserve the restrictions needed for equivalence.
Substitute the expression for \(y'\) into the differentiated equation, and confirm that any transformed relation still respects the original domain.
Check Your Understanding
For each question, pay attention both to the algebra and to the domain where the stated operation is valid.
- For \(\sqrt{2x+y}=x\), what condition on \(x\) and \(y\) allows you to use the usual derivative formula for the radical?
- Differentiate \(\sqrt{x+y}=y\) implicitly where \(x+y>0\), and state the equation before solving for \(y'\).
- Why does differentiating \(\sqrt{y}\) by the usual formula require \(y>0\), even though the expression itself is defined for \(y=0\)?
- What extra sign condition must accompany \(A=B^2\) if it came from \(\sqrt{A}=B\)?
- Give one reason why dividing an implicit equation by an expression could remove valid points from the original relation.
