From an Implicit Slope to a Tangent Line
A derivative can tell us the slope of a curve at a point, even when the curve is given by a relation between \(x\) and \(y\) rather than by a formula \(y=f(x)\). Once that slope and the point are known, the tangent line is determined by the point-slope equation of a line.
The essential steps are to check that the given point lies on the curve, use implicit differentiation to find the slope there, and substitute the point and slope into the line equation. The point check matters: a slope calculated at some other point on the curve does not determine the tangent line being asked for.
What the Tangent-Line Formula Means
Suppose a portion of an implicitly defined curve can be represented near \((x_0,y_0)\) as \(y=g(x)\), where \(g\) is differentiable and \(g(x_0)=y_0\). The derivative \(g'(x_0)\) is the limiting slope of secant lines joining \((x_0,y_0)\) to nearby points on the graph. When this derivative is finite, the line with that slope through \((x_0,y_0)\) is the tangent line.
In an implicit relation, \(y\) varies with \(x\) along the local curve, so implicit differentiation produces a formula for this slope. The earlier material on implicit differentiation gives the rules needed for that calculation. Here the emphasis is on evaluating the derivative at the specified point and using the result to write the line.
A Derivation of the Tangent-Line Equation
The equation follows directly from the meaning of derivative and the equation of a line. Let \(y=g(x)\) be the local graph, with \(g(x_0)=y_0\), and suppose \(g'(x_0)=m\) is finite. The secant slope from the fixed point to a nearby point on the graph is \(\frac{g(x)-y_0}{x-x_0}\). As \(x\) approaches \(x_0\), this slope approaches \(m\). Therefore the tangent line must pass through \((x_0,y_0)\) and have slope \(m\). Every line with slope \(m\) through that point has equation \(y-y_0=m(x-x_0)\), as required.
This also explains why the method does not require a complete solution for \(y\). Implicit differentiation can give the local slope directly from the relation; the tangent-line equation then needs only that slope and the point.
Substitute its coordinates into the original relation and confirm that the equality holds.
Differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\), and solve for \(\frac{dy}{dx}\).
Substitute the point’s coordinates into the derivative formula. The resulting value is \(m\).
Use \(y-y_0=m(x-x_0)\), then verify that the line passes through the given point.
Worked Examples
Worked Example: Tangent to a Circle
Find the tangent line to \(x^2+y^2=25\) at \((3,4)\).
First check the point: \(3^2+4^2=9+16=25\), so \((3,4)\) lies on the circle. Differentiate both sides with respect to \(x\). Since \(y\) depends on \(x\), differentiating \(y^2\) gives \(2y\frac{dy}{dx}\):
Solve for the derivative. Subtract \(2x\) from both sides and divide by \(2y\), which is nonzero at the point:
At \((3,4)\), the slope is \(m=-\frac{3}{4}\). The point-slope equation is
Equivalently, multiplying by \(4\) and rearranging gives \(3x+4y=25\). Check the point in this line equation: \(3(3)+4(4)=9+16=25\). The slope of \(3x+4y=25\), when written as \(y=-\frac{3}{4}x+\frac{25}{4}\), is \(-\frac{3}{4}\), as required.
Worked Example: A Polynomial Relation
Find the tangent line to \(x^2+xy+y^2=7\) at \((1,2)\).
Check that the point is on the curve: \(1^2+(1)(2)+2^2=1+2+4=7\). Differentiate each term with respect to \(x\). In particular, the product rule gives \(\frac{d}{dx}(xy)=y+x\frac{dy}{dx}\), and the chain rule gives \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\):
Collect the terms involving the derivative and solve:
At \((1,2)\), the denominator is \(1+2(2)=5\), so the derivative is defined there. The slope is \(-\frac{2(1)+2}{1+2(2)}=-\frac{4}{5}\). Thus the tangent line is
To check, substitute \((1,2)\) into the simplified equation: \(4(1)+5(2)=4+10=14\). Solving it for \(y\) gives \(y=-\frac{4}{5}x+\frac{14}{5}\), confirming the slope \(-\frac{4}{5}\).
Worked Example: A Relation Not Convenient to Solve for y
Find the tangent line to \(e^y+xy=x^2\) at \((1,0)\). This relation is not conveniently solved for \(y\) by ordinary algebra, but its implicit derivative can still be evaluated at the point.
Check the point: \(e^0+(1)(0)=1\), while \(1^2=1\), so it lies on the curve. Differentiate the relation with respect to \(x\). The chain rule gives \(\frac{d}{dx}(e^y)=e^y\frac{dy}{dx}\), and the product rule gives \(\frac{d}{dx}(xy)=y+x\frac{dy}{dx}\):
Group the derivative terms and solve:
At \((1,0)\), the numerator is \(2(1)-0=2\) and the denominator is \(e^0+1=1+1=2\). Therefore \(m=2/2=1\). The tangent line is
The line contains the point because \(0=1-1\). Its slope is \(1\), matching the derivative at the point. No explicit formula for \(y\) was needed.
A Common Chain-Rule Pitfall
A frequent error is to differentiate a power of \(y\) as if \(y\) were constant. For example, in the polynomial relation above, it would be incorrect to write
That expression omits the rate at which \(y\) changes with \(x\). The chain rule requires a factor of \(\frac{dy}{dx}\). If this factor is omitted, the derivative calculation for \(x^2+xy+y^2=7\) becomes incorrectly
At \((1,2)\), this wrong equation gives \(2+2+\frac{dy}{dx}+4=0\), so it would produce the incorrect slope \(\frac{dy}{dx}=-8\). The correct derivative of \(y^2\) is \(2y\frac{dy}{dx}\); using it gives \(2+2+\frac{dy}{dx}+4\frac{dy}{dx}=0\), hence \(5\frac{dy}{dx}=-4\) and slope \(-\frac{4}{5}\). The incorrect result fails because it treats a varying quantity as constant during differentiation.
Checks Before You Finish
A tangent-line answer should be supported by both the point and the slope. The point must satisfy the original curve relation, and the line must pass through that point. Its slope must agree with the implicit derivative evaluated there. These checks catch different errors: a point-substitution error can select the wrong location, while a differentiation error can give the wrong line direction.
When presenting a result, it is often useful to leave the equation in point-slope form first. This form displays the point and slope used. A simplified equivalent equation is also acceptable, provided it has not changed the line. For instance, multiplying both sides of a line equation by a nonzero constant preserves the same set of points.
The method applies when the required finite derivative exists at the given point. If solving for the derivative leads to a division by zero, that calculation does not provide a finite slope there. Do not force a point-slope equation with a finite \(m\) unless the conditions for that form have been established.
Check Your Understanding
For each problem, identify the point check, the slope, and the resulting tangent-line equation.
- For \(x^2+y^2=13\), find the tangent line at \((2,3)\).
- For \(x^2+xy+y^2=3\), find the tangent line at \((1,1)\).
- For \(e^y+xy=x^2+1\), verify that \((0,0)\) lies on the curve, then find the tangent line there.
- Why does differentiating \(y^3\) with respect to \(x\) produce \(3y^2\frac{dy}{dx}\), rather than \(3y^2\)?
- What two checks can be made directly on a proposed tangent-line equation and its slope?
