Why Higher Derivatives Need Care
First derivatives of implicit and inverse functions often come from one carefully differentiated equation. For second and third derivatives, the same idea still applies, but each new differentiation creates additional terms: derivatives of the branch itself, products of lower derivatives, and denominators that must be checked.
The safest method is to keep the defining equation in view. Differentiate it one step at a time, collect the highest derivative, and only then evaluate at the point of interest. This makes the chain rule visible and gives an equation with which to check the answer.
Conditions for a Local Branch
Suppose a relation is given by \(F(x,y)=0\), and it contains a point \((x_0,y_0)\). A local implicit branch is a function \(y=\phi(x)\), defined near \(x_0\), for which \(\phi(x_0)=y_0\) and \(F(x,\phi(x))=0\).
For an inverse function, suppose \(u=g(x)\), with \(g\) sufficiently differentiable near \(x_0\). If \(g'(x_0)\ne0\), the inverse function theorem gives a local inverse \(x=f(u)\) near \(u_0=g(x_0)\). Higher-derivative formulas for \(f\) divide by powers of \(g'(x_0)\), so this nonzero condition matters in the calculation as well as in the theorem.
Implicit Derivatives: The Equations Behind the Formulas
Along a branch \(y=\phi(x)\), differentiate \(F(x,y)=0\) with respect to \(x\), remembering that \(y\) depends on \(x\). The chain rule gives the first derivative equation. Differentiating again gives the second derivative equation:
Here the partial derivatives of \(F\) are evaluated at \((x,y(x))\). If \(F_y\ne0\), the second equation can be solved for \(y''\). A third differentiation gives:
These equations are useful as working formulas, but it is important to understand why the coefficients occur. For example, differentiating \(F_x\) along the branch gives \(F_{xx}+F_{xy}y'\), while differentiating \(F_y y'\) gives \((F_{yx}+F_{yy}y')y'+F_y y''\). Since \(F_{xy}=F_{yx}\) when the second partial derivatives are continuous, adding the terms produces \(2F_{xy}y'\) in the second derivative equation. Further differentiation and the product rule give the third derivative equation.
Worked Example: Third Derivative on a Circle
Worked Example: Third Derivative on a Circle
Find \(y'\), \(y''\), and \(y'''\) at \((3,4)\) on the local upper branch of \(x^2+y^2=25\).
Let \(F(x,y)=x^2+y^2-25\). At the given point, \(F(3,4)=9+16-25=0\), and \(F_y=2y\), so \(F_y(3,4)=8\ne0\). Thus the relation defines a smooth local branch there.
Differentiate the relation once:
At \((3,4)\), this gives \(3+4y'=0\), so \(y'=-3/4\).
Differentiate \(x+yy'=0\) again:
Substitute \(x=3\), \(y=4\), and \(y'=-3/4\):
Differentiate the second-derivative equation. The derivative of \((y')^2\) is \(2y'y''\), and the derivative of \(yy''\) is \(y'y''+yy'''\). Therefore:
At the point, \(y'y''=(-3/4)(-25/64)=75/256\). Hence
Check: The first differentiated equation gives \(3+4(-3/4)=0\). The second gives \(1+9/16+4(-25/64)=0\). The third gives \(3(75/256)+4(-225/1024)=225/256-225/256=0\). Each value satisfies its corresponding differentiated relation.
Derivatives of an Inverse Function
Let \(f\) be the local inverse of \(g\), so \(g(f(u))=u\). Differentiate this identity rather than trying to manipulate an inverse formula directly. The chain rule first gives \(g'(f(u))f'(u)=1\). Differentiating twice more gives the equations from which the second and third derivatives follow.
To see the last two formulas, differentiate \(g(f(u))=u\):
The second equation and \(f'=1/g'(f)\) give \(f''=-g''(f)/(g'(f))^3\). One more differentiation of the second equation yields
Substitute the expressions for \(f'\) and \(f''\), then divide by \(g'(f)\ne0\). The result is \(f'''=3(g''(f))^2/(g'(f))^5-g'''(f)/(g'(f))^4\), as stated. The condition \(g'\ne0\) is necessary for these divisions and for the local inverse theorem.
Worked Example: Third Derivative of a Polynomial Inverse
Worked Example: Third Derivative of a Polynomial Inverse
Let \(g(x)=x+x^3\), and let \(f\) be its local inverse. Find \(f'(2)\), \(f''(2)\), and \(f'''(2)\).
First identify the corresponding input \(x\): \(g(1)=1+1=2\), so \(f(2)=1\). At \(x=1\),
Since \(g'(1)=4\ne0\), the inverse is differentiable near \(u=2\), and the formulas apply. They give
Check: The twice-differentiated identity requires \(g''(f)(f')^2+g'(f)f''=0\). At \(u=2\), its left side is \(6(1/4)^2+4(-3/32)=3/8-3/8=0\). The third-derivative identity gives \(6(1/4)^3+18(1/4)(-3/32)+4(21/256)=3/32-27/64+21/64=0\). Both checks confirm the computed derivatives.
Worked Example: A Third Derivative from an Implicit Relation
Worked Example: A Third Derivative from an Implicit Relation
For the branch through \((1,1)\) defined by \(y^2+xy-2x=0\), find \(y'\), \(y''\), and \(y'''\) at \(x=1\).
The point lies on the curve because \(1+1-2=0\). For \(F(x,y)=y^2+xy-2x\), \(F_y=2y+x\), so \(F_y(1,1)=3\ne0\). The implicit function theorem therefore gives a smooth local branch through the point.
Differentiate the equation:
At \((1,1)\), \(3y'=1\), so \(y'=1/3\).
Differentiate the first differentiated equation in its expanded form:
At the point, this becomes \(2/9+3y''+2/3=0\). Thus \(3y''=-8/9\) and \(y''=-8/27\).
Differentiate once more. Applying the product rule to \((2y+x)y''\) gives \((2y'+1)y''+(2y+x)y'''\), so
At \((1,1)\), this is \(5(-8/27)+3y'''=0\), giving \(y'''=40/81\).
Check: The first derivative equation evaluates to \(2(1)(1/3)+1+1/3-2=0\). The second evaluates to \(2/9+3(-8/27)+2/3=0\). The third evaluates to \(5(-8/27)+3(40/81)=-40/27+40/27=0\). The point and all three derivative values satisfy the equations used to obtain them.
A Common Mistake: Losing a Chain-Rule Term
When differentiating \(x+yy'=0\), a common incorrect step is to write \(1+yy''=0\), as if \(y\) were constant in the product \(yy'\). That omits the derivative of the first factor. The correct product rule is
so the full equation is \(1+(y')^2+yy''=0\). On the circle at \((3,4)\), the incorrect calculation would give \(y''=-1/4\), but substitution into the correct equation gives \(1+9/16+4(-1/4)=9/16\ne0\). Including the missing term instead gives \(y''=-25/64\), for which \(1+9/16+4(-25/64)=0\).
A Reliable Calculation Strategy
For either kind of problem, the calculation is more dependable when organized around the defining identity and its hypotheses:
Substitute the given coordinates into the defining relation, or check that the inverse input \(u\) equals \(g(x)\).
For an implicit branch, evaluate \(F_y\). For an inverse, evaluate \(g'\). Confirm that the relevant value is nonzero before dividing.
Write each differentiated equation before substituting point values, so that product-rule and chain-rule terms remain visible.
Isolate the highest derivative, evaluate it, and substitute the answer into the differentiated equation to test it.
Higher derivatives describe how a slope itself changes and how that change continues to evolve. In implicit and inverse problems, their values depend not only on the original equation or function but also on the branch and the point. Checking the defining relation, the nonzero denominator, and each differentiated identity keeps those dependencies explicit.
Check Your Understanding
For each question, identify the defining equation and the condition that permits the required division.
- For \(F(x,y)=0\), write the first- and second-derivative equations along a branch \(y=\phi(x)\).
- Why must \(F_y(x_0,y_0)\ne0\) be checked before solving the implicit second-derivative equation for \(y''\)?
- If \(f\) is the local inverse of \(g\), what identity should be differentiated to obtain derivatives of \(f\)?
- For \(g(x)=x+x^3\), what value of \(u\) corresponds to \(x=1\), and what condition verifies that the local inverse formulas apply there?
- When differentiating a product \(yy'\), which two terms must appear?
