Why a Second Quotient Derivative Needs Care
When a function is a quotient, its first derivative already contains several terms. Finding the second derivative means differentiating that entire expression: both its numerator and its denominator vary with \(x\). Keeping track of those changes is the main challenge.
A reliable approach is to write the first derivative as a single quotient, name its numerator, and then apply the quotient rule again. Simplification can make the work shorter, but only after the derivative has been formed correctly.
Conditions and a Useful Formula
Suppose \(u\) and \(v\) are twice differentiable on an open interval, and \(v(x)\ne 0\) throughout that interval. Then \(f(x)=\frac{u(x)}{v(x)}\) is defined there, and its second derivative can be found by differentiating the quotient-rule expression for \(f'(x)\).
The formula is useful when the derivatives of \(u\) and \(v\) are easy to identify. In actual calculations, it is often safer to first differentiate the quotient for \(f'\), then simplify. That method makes it easier to see where each term comes from and to catch an algebra error.
Why the Formula Works
Start from the first derivative and call its numerator \(N(x)\). The quotient rule gives \(f'(x)=\frac{N(x)}{v(x)^2}\), where \(N(x)=u'(x)v(x)-u(x)v'(x)\). Differentiate \(N\) using the product rule on each product:
The two middle terms cancel. Now apply the quotient rule to \(N(x)/v(x)^2\). Differentiating the denominator uses the chain rule: \(\frac{d}{dx}(v(x)^2)=2v(x)v'(x)\). Therefore,
Expanding the numerator and dividing each term by \(v(x)^3\) yields the stated formula. This derivation also explains why the terms involving \(u'v'\) and \(u(v')^2\) appear: they come from differentiating the denominator \(v^2\) in the first-derivative expression.
A Reliable Calculation Process
Find where the original denominator is nonzero. The second derivative formula is valid on intervals where the denominator stays nonzero and both numerator and denominator are twice differentiable.
Use the quotient rule and simplify its numerator carefully.
Apply the quotient rule again, including the derivative of the squared denominator.
Combine terms only when useful, then substitute a convenient input into both the derived expression and an equivalent formula or unsimplified calculation.
Worked Examples
Worked Example: A Linear Numerator over a Quadratic Denominator
Find the second derivative of \(f(x)=\frac{3x+1}{x^2+1}\), then check it at \(x=0\) and \(x=1\). The denominator \(x^2+1\) is positive for every real \(x\), so the function and its derivatives are defined for all real \(x\).
Let \(u=3x+1\) and \(v=x^2+1\). The first derivatives are \(u'=3\) and \(v'=2x\). Applying the quotient rule,
Set \(A(x)=-3x^2-2x+3\), so \(A'(x)=-6x-2\). Differentiating \(A(x)/(x^2+1)^2\), the derivative of the denominator is \(4x(x^2+1)\). Thus
For a check at \(x=0\), substitute the values \(u=1\), \(v=1\), \(u'=3\), \(v'=0\), \(u''=0\), and \(v''=2\) into the general formula:
The simplified expression gives \(\frac{-2}{1}=-2\), as required. At \(x=1\), its numerator is \(6+6-18-2=-8\), and its denominator is \((1+1)^3=8\), so \(f''(1)=-1\). The general formula gives \(u=4\), \(v=2\), \(u'=3\), \(v'=2\), \(u''=0\), and \(v''=2\), hence \(0-\frac{8}{4}-\frac{12}{4}+\frac{32}{8}=-1\). The two checks agree.
Worked Example: A Quadratic Numerator over a Linear Denominator
Find the second derivative of \(g(x)=\frac{x^2+2}{x-1}\), and check it at \(x=2\). The function is defined only when \(x\ne1\); the calculation applies on either interval not containing \(1\).
Take \(u=x^2+2\) and \(v=x-1\). Then \(u'=2x\) and \(v'=1\). The first derivative is
Let \(B(x)=x^2-2x-2\), so \(B'(x)=2x-2\). Differentiating \(B(x)/(x-1)^2\) gives
At \(x=2\), this expression gives \(g''(2)=6\). Check using the general formula: \(u=6\), \(v=1\), \(u'=4\), \(v'=1\), \(u''=2\), and \(v''=0\). Therefore,
This confirms the simplified result at the selected input.
Worked Example: A Quotient with a Simple First Derivative
Find the second derivative of \(h(x)=\frac{x^2}{x^2+1}\), and check the result at \(x=0\) and \(x=1\). Since \(x^2+1>0\), the function is defined for every real \(x\).
The quotient rule gives
Differentiate this quotient. The numerator has derivative \(2\), and the denominator has derivative \(4x(x^2+1)\):
At \(x=0\), the simplified expression gives \(h''(0)=2\). The general formula also gives \(u=0\), \(v=1\), \(u'=0\), \(v'=0\), \(u''=2\), and \(v''=2\), so \(h''(0)=2\). At \(x=1\), the simplified expression gives \(\frac{2-6}{8}=-\frac12\). The formula gives
Both selected-input checks agree with the derived expression.
Common Pitfall: Differentiating the Denominator Incorrectly
A frequent error is to differentiate \(v(x)^2\) as though \(v\) were just \(x\). The chain rule requires the additional factor \(v'(x)\): \(\frac{d}{dx}[v(x)^2]=2v(x)v'(x)\).
For example, in the second example the first derivative is \(\frac{x^2-2x-2}{(x-1)^2}\). A wrong calculation might say that the denominator's derivative is \(2(x-1)\), then write
This happens to show the right denominator derivative for this particular \(v=x-1\), because \(v'=1\). But the shortcut is not generally valid. If \(v=x^2+1\), its squared denominator differentiates to \(4x(x^2+1)\), not \(2(x^2+1)\). Omitting the factor \(v'=2x\) changes the result. Always differentiate the entire denominator as a composite expression.
When to Simplify and How to Check
Simplifying the first derivative before differentiating can reduce the number of terms, as in \(h'(x)=\frac{2x}{(x^2+1)^2}\). However, canceling factors is legitimate only when the algebraic factorization is correct and the original domain restrictions are retained. For instance, a formula obtained after cancellation does not make an original function defined at a point where its denominator was zero.
A practical check is to choose an input in the domain and evaluate the result in two ways: substitute into the simplified second derivative, and substitute into the general formula using \(u\), \(v\), and their derivatives. Agreement does not replace the derivation, but it can reveal sign, expansion, or denominator errors.
Check Your Understanding
For each question, show enough work to make clear how the quotient-rule expression is differentiated.
- For \(f(x)=\frac{x+2}{x^2+1}\), identify the domain and find \(f'(x)\).
- For the same function, find \(f''(x)\), leaving your result as a single rational expression.
- State the derivative of \(v(x)^2\) when \(v(x)=x^2-3\), and explain which factor is missed by the incorrect expression \(2v(x)\).
- Use the general second-derivative formula to check the value of \(f''(0)\) for \(f(x)=\frac{x^2}{x^2+1}\).
- Why must a point where the original denominator is zero remain excluded, even if algebraic simplification appears to remove a factor?
