A Diagnostic Routine for Implicit Differentiation
Implicit differentiation problems can involve several issues at once. An equation may combine products, composite functions, or restricted domains; a proposed point may not lie on the curve; and the derivative formula may have a zero denominator. A reliable solution therefore does more than differentiate and substitute. It checks whether the setup is valid, solves the derivative equation carefully, and interprets what the result says about the tangent.
For a relation written as \(F(x,y)=0\), treat \(y\) as a function of \(x\) when differentiating along a curve. Then use the resulting equation to find \(\frac{dy}{dx}\), provided the coefficient of \(\frac{dy}{dx}\) is nonzero. If that coefficient is zero, do not divide automatically: the point may have a vertical tangent or may be singular.
Why Differentiating the Relation Works
Suppose a differentiable branch of the curve can be written locally as \(y=y(x)\). Along this branch, the expression \(F(x,y(x))\) equals zero for every \(x\) in a suitable interval. Since it is a constant function, its derivative is zero. The chain rule gives the key equation.
Here \(F_x\) and \(F_y\) are the partial derivatives of \(F\), found by differentiating with respect to one variable while holding the other fixed. If \(F_y(a,b)\ne 0\), we can solve this equation for the slope at \((a,b)\).
The reason for the condition is visible in the differentiated equation: \(F_y\) is the coefficient of \(y'\). Dividing by zero is not permitted, and the remaining equation must instead be interpreted. For a regular point, where at least one of \(F_x\) and \(F_y\) is nonzero, the tangent direction \((dx,dy)\) satisfies
If \(F_y\ne 0\), this direction has a finite slope \(dy/dx=-F_x/F_y\). If \(F_y=0\) but \(F_x\ne 0\), then \(dx=0\) for a tangent direction, so the tangent is vertical. If both partial derivatives are zero, this first-derivative test is inconclusive.
A Practical Order of Checks
Substitute both coordinates into the original relation. Confirm that the equation is satisfied.
Verify restrictions such as positive logarithm arguments and nonnegative radicands. For ordinary differentiation formulas, also check whether the point is in the interior of the allowed domain.
Use the product rule for products and the chain rule whenever \(y\) occurs inside a function of \(y\). Include a factor of \(y'\) when differentiating with respect to \(x\).
Solve the resulting equation only after collecting terms, and check whether the coefficient of \(y'\) is zero at the point.
A finite slope is an ordinary tangent slope; slope zero means horizontal. A vertical tangent requires a regular point with \(F_y=0\) and \(F_x\ne0\). If both partials vanish, more analysis is needed.
Worked Examples
Worked Example: A Polynomial Relation and a Point Check
Find and interpret the slope of \(x^2+xy+y^2=7\) at \((1,2)\). First check membership:
The point is on the relation, and this polynomial has no domain restrictions. Differentiate each term. The product rule gives \(\frac{d}{dx}(xy)=y+xy'\), while \(\frac{d}{dx}(y^2)=2yy'\).
Collect the derivative terms and solve:
At \((1,2)\), the numerator is \(2(1)+2=4\), and the denominator is \(1+2(2)=5\). Since the denominator is nonzero, the slope is finite:
The tangent line through \((1,2)\) is \(y-2=-\frac45(x-1)\). Substituting the point into this line gives \(2-2=0\), as required. The slope calculation is also checked directly in the differentiated equation: \(2(1)+2+(1)(-\frac45)+2(2)(-\frac45)=4-\frac45-\frac{16}{5}=0\).
For contrast, \((1,1)\) is not a valid point for this calculation: \(1^2+(1)(1)+1^2=3\), not \(7\). A slope formula evaluated at a point off the relation does not give a tangent slope to this curve.
Worked Example: A Radical and a Logarithm
Consider \(\sqrt{y+1}+\ln x=2\) at \((1,3)\). The domain requires \(y+1\ge 0\) and \(x>0\); the point satisfies both, and it is in the interior of these restrictions. It also lies on the relation because \(\sqrt{3+1}+\ln 1=2+0=2\).
Differentiate implicitly. The chain rule gives \(\frac{d}{dx}\sqrt{y+1}=\frac{y'}{2\sqrt{y+1}}\), and \(\frac{d}{dx}\ln x=1/x\). Thus
Solving and then substituting the point gives
The coefficient of \(y'\) in the differentiated equation is \(1/(2\sqrt{y+1})\), which equals \(1/4\) at the point and is nonzero. So the finite slope is valid. The tangent line is \(y-3=-4(x-1)\). To check the derivative, substitute \(x=1\), \(y=3\), and \(y'=-4\) into the differentiated equation: \(-4/4+1=0\).
Worked Example: A Product and an Exponential Term
Find the slope on \(e^y+xy=e+1\) at \((1,1)\). The point is valid because \(e^1+(1)(1)=e+1\), and exponential and product expressions are defined for all real \(x,y\).
Differentiate \(e^y\) by the chain rule and \(xy\) by the product rule:
Collecting \(y'\) yields \((e^y+x)y'=-y\). At \((1,1)\), the coefficient is \(e+1\), which is nonzero. Therefore
Substitution into the differentiated equation checks the answer: \(e(-1/(e+1))+1+1(-1/(e+1))=-(e+1)/(e+1)+1=0\). The slope is finite and nonzero, so the tangent is neither horizontal nor vertical.
Worked Example: Vertical Tangent Versus Singular Point
For \(x^2+y^2=25\) at \((5,0)\), membership holds because \(25+0=25\). Differentiation gives \(2x+2yy'=0\), so the coefficient of \(y'\) is \(2y=0\) at the point. The slope quotient is not available there. But \(F_x=2x=10\ne0\), so this is a regular point, not a singular one. The tangent-direction equation is \(10\,dx+0\,dy=0\), which requires \(dx=0\): the tangent is vertical. Indeed, the circle has tangent line \(x=5\) there.
Now consider \(x^2-y^2=0\) at \((0,0)\). Membership holds, but \(F_x=2x=0\) and \(F_y=-2y=0\). Differentiation gives \(2x-2yy'=0\), which becomes \(0=0\) at the origin and determines no slope. The relation factors as \((x-y)(x+y)=0\), so it consists of the two lines \(y=x\) and \(y=-x\). At the origin those branches have slopes \(1\) and \(-1\). Thus the point is singular and has no unique tangent direction. The first-derivative test alone cannot classify it.
Common Mistake: Losing a Derivative Term
A frequent error is to differentiate a product containing \(y\) as though \(y\) were constant. In the equation \(e^y+xy=e+1\), a mistaken calculation might write \(\frac{d}{dx}(xy)=y\), leading to \(e^y y'+y=0\). At \((1,1)\), this gives the incorrect slope \(-1/e\).
The product rule actually gives \(\frac{d}{dx}(xy)=y+xy'\). The correct equation is \(e^y y'+y+xy'=0\), so the slope at \((1,1)\) is \(-1/(e+1)\), not \(-1/e\). Substitution verifies the correction: \(e(-1/(e+1))+1-1/(e+1)=0\). The error is not a small arithmetic slip; it removes a term that changes the derivative equation.
What the Diagnosis Tells You
A correct numerical value for \(y'\) is only one possible outcome. A finite value gives the slope of a local graph \(y(x)\) under the stated regularity conditions. A zero value gives a horizontal tangent. At a regular point with \(F_y=0\) and \(F_x\ne0\), the tangent is vertical rather than absent. When both partial derivatives vanish, the first-derivative equation cannot settle the question: the relation may have multiple branches, a single tangent determined by higher-order behavior, or another singular feature.
Domain and membership checks are equally important. A point outside the curve has no tangent to that curve, and a point where a logarithm or radical is not defined cannot be rescued by a formal derivative calculation. At a domain boundary, the usual local differentiability conclusions may also fail. Keeping these checks separate from the algebra makes it easier to locate exactly what a problem does—and does not—establish.
Check Your Understanding
For each question, identify the relevant checks before deciding what the derivative calculation establishes.
- For \(x^2+xy=6\), does \((2,1)\) lie on the relation? If so, find the slope there and check it in the differentiated equation.
- For \(\sqrt{y}+\ln x=3\), list the domain restrictions and explain why a point with \(x\le 0\) cannot be used.
- When differentiating \(x\sin y\), what terms result, and where does \(y'\) appear?
- For a relation \(F(x,y)=0\), what conditions distinguish a regular vertical tangent from a point where the first-derivative test is inconclusive?
- Why does \(F_y=0\) by itself not prove that a point is singular?
