Deriving the Reciprocal Rule from the Chain Rule

A Reciprocal Is a Composition

Many functions have the form \(\dfrac{1}{u(x)}\) with a numerator that is just the constant \(1\): \(\dfrac{1}{x^2+1}\), \(\dfrac{1}{3x-5}\), \(\dfrac{1}{\sin x}\). The Quotient Rule would handle them, but there is a shorter route that also explains why the answer has the shape it does. The key observation is that \(\dfrac{1}{u(x)}\) is a composition. Let \(r(t)=\dfrac1t\) be the reciprocal function, defined for \(t\ne0\). Then \(\dfrac{1}{u(x)}=r\bigl(u(x)\bigr)\): first apply \(u\), then take the reciprocal of the result. Because the Chain Rule differentiates compositions, all that is needed is the derivative of \(r\).

The Derivative of the Reciprocal Function

Fix \(t\ne0\) and use the limit definition. For \(h\) small enough that \(t+h\ne0\),

As \(h\to0\), the denominator \(t(t+h)\) tends to \(t^2\ne0\), so the quotient tends to \(-\dfrac{1}{t^2}\). Hence \(r'(t)=-\dfrac1{t^2}\) for every \(t\ne0\). This agrees with the Power Rule applied to \(t^{-1}\), which gives \(-1\cdot t^{-2}\), but the computation above does not depend on the Power Rule for negative exponents.

The Reciprocal Rule

Derivation. Write \(\dfrac1u=r\circ u\). The inner function \(u\) is differentiable at \(x\), and the outer function \(r\) is differentiable at the point \(u(x)\) because \(u(x)\ne0\). The Chain Rule then says that \((r\circ u)'(x)=r'\bigl(u(x)\bigr)\,u'(x)\). Substituting \(r'(t)=-1/t^2\) with \(t=u(x)\) gives

The condition \(u(x)\ne0\) is essential. At a point where \(u(x)=0\), the function \(\dfrac1u\) is not even defined, so there is nothing to differentiate. Notice also the structure of the result: the denominator is the square of the original denominator, and the numerator is the negative of the derivative of the original denominator.

Consistency with the Quotient Rule

The Reciprocal Rule is the special case of the Quotient Rule with numerator \(1\). Take \(\dfrac{N}{D}\) with \(N=1\) and \(D=u\). The Quotient Rule gives \(\dfrac{N'D-ND'}{D^2}=\dfrac{0\cdot u-1\cdot u'}{u^2}=-\dfrac{u'}{u^2}\), the same expression. The advantage of the reciprocal form is speed: when the numerator is constant there is one term to compute instead of two. A constant numerator \(k\) is handled by pulling out the constant: \(\dfrac{d}{dx}\dfrac{k}{u}=-\dfrac{k\,u'}{u^2}\).

Common Mistakes

Dropping the inner derivative. A frequent error is to differentiate only the reciprocal and forget the Chain Rule factor. For \(f(x)=\dfrac{1}{x^2+1}\), writing \(f'(x)=-\dfrac{1}{(x^2+1)^2}\) is wrong. At \(x=1\) it gives \(-\dfrac14\), but the correct value is \(-\dfrac12\); the missing factor is \(u'(1)=2\).

Reciprocating the derivative. The derivative of \(\dfrac1u\) is not \(\dfrac1{u'}\). For \(u=3x-5\), \(\dfrac1{u'}=\dfrac13\) is a constant, whereas \(\dfrac1u\) clearly changes with \(x\), so its derivative cannot be constant.

Losing the sign. The reciprocal function is decreasing wherever it is defined, so an increasing \(u\) (positive \(u'\)) makes \(\dfrac1u\) decrease, and the derivative must be negative. In the linear example, \(u'=3>0\) and \(g'<0\), as the formula says.

Summary of the Rule

To use the rule well, identify the denominator, differentiate it, and place its derivative over the square of the denominator with a minus sign. Then simplify. Two quick sanity checks catch most errors. First, the sign: a denominator that is increasing at the point forces a negative derivative. Second, the size: if the denominator is large in absolute value, the reciprocal is small and nearly flat, so the derivative should be small as well, which the squared denominator guarantees.

Reciprocals show up constantly in later work. The derivative of \(\sec x\) and \(\csc x\), the derivative of a rational function with a constant numerator, and the derivative of an inverse function evaluated by \(\dfrac{1}{f'(f^{-1}(y))}\) all use the same pattern: a derivative divided by a square, with a minus sign. A reader who understands where the minus sign and the square come from, namely from \(r'(t)=-1/t^2\), can reconstruct each of these results instead of memorizing them. When a problem seems to need the Quotient Rule but the numerator is constant, use the Reciprocal Rule first, because it is shorter and leaves less room for algebra errors.

Where This Fits in the Course

This tutorial belongs to the module “Product, Quotient, and Chain Rules”. The tutorial just before it is “Quotient Rule for Polynomial Functions,” and the one that follows is “Simplifying Rational Functions Before Differentiation.”

Powers of a Reciprocal

The same idea extends to \(\dfrac{1}{u^n}\) for a positive integer \(n\). Write \(\dfrac1{u^n}=r_n(u)\) with \(r_n(t)=t^{-n}\). The Power Rule gives \(r_n'(t)=-n\,t^{-n-1}\), and the Chain Rule then yields

For \(n=1\) this is the Reciprocal Rule. For example, with \(u=x^2+1\) and \(n=3\), \(\dfrac{d}{dx}\dfrac1{(x^2+1)^3}=-\dfrac{3\cdot2x}{(x^2+1)^4}=-\dfrac{6x}{(x^2+1)^4}\), and at \(x=1\) this equals \(-\dfrac{6}{16}=-0.375\). A difference quotient of \(\dfrac1{(x^2+1)^3}\) near \(x=1\) confirms the value.

Practice Problems with Answers

Try each problem before reading the answer. Each answer below was verified by differentiation or by evaluating a difference quotient or integral numerically.

1. Differentiate \(\dfrac{1}{x^3+2x}\) and evaluate at \(x=1\).
   Answer. \(-\dfrac{3x^2+2}{(x^3+2x)^2}\); at \(x=1\) this is \(-\dfrac59\).
2. Differentiate \(\dfrac{4}{2x+1}\).
   Answer. \(-\dfrac{8}{(2x+1)^2}\); at \(x=0\) the value is \(-8\).
3. Derive \((\sec x)'\) from \(\sec x=\dfrac1{\cos x}\).
   Answer. \(-\dfrac{-\sin x}{\cos^2x}=\dfrac{\sin x}{\cos^2x}=\sec x\tan x\).
4. Where is \(\dfrac1{x^2-4}\) differentiable, and what is the derivative at \(x=3\)?
   Answer. For \(x\ne\pm2\), with derivative \(-\dfrac{2x}{(x^2-4)^2}\); at \(x=3\) it equals \(-\dfrac{6}{25}\).