A Mixed Approach to Implicit Differentiation
In an implicit equation, \(x\) and \(y\) are related without \(y\) being isolated as a formula in \(x\). When a curve contains polynomial, exponential, logarithmic, or trigonometric terms, the central task remains the same: differentiate the relation while treating \(y\) as a function of \(x\). The chain rule then supplies a factor of \(y'\) whenever a differentiated expression depends on \(y\).
The calculations in this workshop combine several standard tasks: finding a slope at a point, writing the tangent line there, and finding a second derivative. Each answer should be checked against the differentiated relation. For a curve written as \(F(x,y)=0\), a specified point must also satisfy the original equation before its slope or tangent line is calculated.
A General Formula to Organize the Work
Suppose a curve is given by \(F(x,y)=0\), and suppose a twice-differentiable local branch \(y=y(x)\) passes through the point under consideration. The notation \(F_x\) and \(F_y\) denotes partial derivatives with respect to \(x\) and \(y\), respectively. Differentiating the relation along the branch gives
If \(F_y\ne 0\) at the point, this equation determines the slope there as \(y'=-F_x/F_y\). If \(F_y=0\), this division is not allowed; the equation may indicate a vertical tangent or a failure of a differentiable local branch, and the point requires separate analysis.
When \(F\) has continuous second partial derivatives, differentiate the first-derivative equation once more. The derivative of \(F_x\) along the branch is \(F_{xx}+F_{xy}y'\). The derivative of \(F_y y'\) is \((F_{yx}+F_{yy}y')y'+F_y y''\). Since \(F_{xy}=F_{yx}\), collecting terms gives
Thus, when \(F_y\ne 0\), the second derivative can be found by solving this equation for \(y''\). This formula is a bookkeeping aid, not a replacement for careful differentiation: in a particular problem it is often clearer to differentiate the first-derivative equation directly.
Worked Example: A Polynomial Curve and Its Tangent Line
Worked Example: A Polynomial Curve and Its Tangent Line
Find the tangent line to \(x^2+xy+y^2=7\) at \((1,2)\).
Step 1: Check the point. Substitute \(x=1\) and \(y=2\) into the curve equation:
The point lies on the curve.
Step 2: Differentiate implicitly. Differentiate each term with respect to \(x\), using the product rule on \(xy\) and the chain rule on \(y^2\):
Collect the terms containing \(y'\):
At \((1,2)\), the denominator is \(1+2(2)=5\), so it is nonzero. The slope is
Step 3: Write and check the tangent line. The point-slope equation is
To check the slope, substitute \((1,2)\) and \(y'=-4/5\) into the differentiated relation. Its left side is \(2(1)+2+(1+2(2))(-4/5)=4-4=0\), as required. The line also passes through \((1,2)\), so it has the required point and slope.
Worked Example: A Polynomial Curve and a Second Derivative
Worked Example: A Polynomial Curve and a Second Derivative
For \(x^2+y^2=25\), find \(y'\) and \(y''\) at \((3,4)\).
Step 1: Check the point and find \(y'\).
Differentiate the relation:
Since \(y=4\ne0\) at the point, this gives \(y'=-3/4\).
Step 2: Differentiate again. Differentiate \(2x+2yy'=0\), applying the product rule to \(yy'\):
At \((3,4)\), use \(y'=-3/4\):
Check. Substituting the computed values into the second differentiated equation gives \(2+2(9/16+4(-25/64))=2+2(9/16-25/16)=2-2=0\). The second derivative satisfies the equation at the point.
Worked Example: An Exponential Relation
Worked Example: An Exponential Relation
For \(e^y+xy=1+x\), find \(y'\) and \(y''\) at \((0,0)\).
Step 1: Check the point. The left side is \(e^0+(0)(0)=1\), and the right side is \(1+0=1\).
Step 2: Find the first derivative. The chain rule gives \(\frac{d}{dx}e^y=e^y y'\), while the product rule gives \(\frac{d}{dx}(xy)=y+xy'\). Therefore,
At \((0,0)\), this becomes \(y'+0+0=1\), so \(y'=1\).
Step 3: Find the second derivative. Differentiate the entire first-derivative equation. In particular, \(\frac{d}{dx}(e^y y')=e^y(y')^2+e^y y''\), and \(\frac{d}{dx}(xy')=y'+xy''\). Hence
At \((0,0)\), use \(e^y=1\) and \(y'=1\):
Check. Substitution in the second-derivative equation gives \(1+(-3)+1+1+0=0\). The result satisfies the differentiated relation.
Worked Example: A Logarithmic Relation
Worked Example: A Logarithmic Relation
For \(\ln(xy)=x-y\), find the tangent line and the second derivative at \((1,1)\).
Step 1: Check the point and domain. Since \(xy=1>0\), the logarithm is defined near this point. Also, \(\ln(1)=0=1-1\), so the point lies on the curve.
Step 2: Find the first derivative. Differentiating \(\ln(xy)\) gives \(\frac{x y'+y}{xy}=\frac1x+\frac{y'}y\). Thus
At \((1,1)\), this becomes \(1+y'=1-y'\), so \(2y'=0\) and \(y'=0\). The tangent line is therefore \(y-1=0(x-1)\), or \(y=1\).
Step 3: Find the second derivative. Differentiate the first-derivative equation:
At \((1,1)\), with \(y'=0\), this gives \(-1+y''=-y''\). Therefore \(2y''=1\), so \(y''=1/2\).
Check. In the differentiated equation, substitution gives \(-1+1/2-0=-1/2\), equal to the right side \(-y''=-1/2\). The tangent line passes through \((1,1)\) and has the computed slope zero.
Worked Example: A Trigonometric Relation
Worked Example: A Trigonometric Relation
For \(\sin x+\sin y=1\), find the tangent line and \(y''\) at \((\pi/6,\pi/6)\).
Step 1: Check the point.
Step 2: Find the slope. Differentiation gives
At the specified point, both cosine values are \(\sqrt{3}/2\). Therefore \(\sqrt{3}/2+(\sqrt{3}/2)y'=0\), which gives \(y'=-1\). The tangent line is
Step 3: Find the second derivative. Differentiate the slope equation, using the chain rule on \(\cos y\):
At the point, \(\sin x=\sin y=1/2\), \(\cos y=\sqrt{3}/2\), and \((y')^2=1\). Thus
Check. Substituting \(y''=2/\sqrt{3}\) into the second-derivative equation gives \(-1/2-1/2+(\sqrt{3}/2)(2/\sqrt{3})=-1+1=0\). The first-derivative equation also gives \(\sqrt{3}/2-(\sqrt{3}/2)=0\) at the point.
A Common Chain-Rule Error
A frequent mistake is differentiating a function of \(y\) as though \(y\) were a constant. For example, the incorrect calculation \(\frac{d}{dx}e^y=e^y\) omits the factor \(y'\). The correct calculation is \(\frac{d}{dx}e^y=e^y y'\), because \(y\) depends on \(x\).
The error is visible in the exponential example. If one incorrectly differentiates \(e^y+xy=1+x\) as \(e^y+y+xy'=1\), then evaluating at \((0,0)\) gives \(1=1\), which supplies no slope at all. With the chain-rule factor included, the equation is \(e^y y'+y+xy'=1\), and at the point it gives \(y'=1\). Omitting the factor has not produced a different valid method; it has lost the information needed to find the derivative.
A Reliable Sequence of Checks
For a mixed implicit-differentiation problem, keep the algebra organized by following the calculation in this order:
Substitute its coordinates into the original relation. A point not on the curve cannot be the point of tangency for that curve.
Use product and chain rules wherever needed, keeping \(y'\) terms visible.
Substitute the point into the derivative equation and check that any denominator used to solve for the derivative is nonzero.
For a tangent line, verify the point and slope. For a second derivative, substitute the values back into the second differentiated equation.
The variety of functions changes the chain-rule details, but not the underlying logic: a point on the curve determines which derivative equation to evaluate, and the derivative equation determines the local slope when its coefficient permits solving for \(y'\). Repeating the process yields a second derivative when the relation is sufficiently differentiable and the branch is well-defined.
Check Your Understanding
For each problem, show the differentiation steps and check the final result in the appropriate equation.
- For \(x^2+xy=6\), verify that \((2,1)\) lies on the curve and find its slope.
- For \(e^y+x=2\), find \(y'\) at \((1,0)\).
- For \(\ln(x+y)=y\), explain why \((1,0)\) lies on the curve, then find the slope there if it exists.
- For \(\sin x+y^2=1\), find the slope at \((0,1)\).
- State the condition that allows the formula \(y'=-F_x/F_y\) to be used at a point on \(F(x,y)=0\).
