Quotient rule

Reciprocal rule

The reciprocal rule is a special case of the quotient rule in which the numerator ⁠${\displaystyle f(x)=1}$⁠. Applying the quotient rule gives $${\displaystyle h^{\prime }(x)=\frac{d}{dx}\left[\frac{1}{g(x)}\right]=\frac{0\cdot g(x)-1\cdot g^{\prime }(x)}{g(x{)}^2}=\frac{-g^{\prime }(x)}{g(x{)}^2}.}$$

Utilizing the chain rule yields the same result.

Proofs

Proof from derivative definition and limit properties

Let ⁠${\displaystyle h(x)=\frac{f(x)}{g(x)}}$⁠. Applying the definition of the derivative and properties of limits gives the following proof, with the term ${\displaystyle f(x)g(x)}$ added and subtracted to allow splitting and factoring in subsequent steps without affecting the value: $${\displaystyle \begin{array}{cc}{h}^{\prime }(x)& =\underset{k\to 0}{lim}\frac{h(x+k)-h(x)}{k}\\ & =\underset{k\to 0}{lim}\frac{\frac{f(x+k)}{g(x+k)}-\frac{f(x)}{g(x)}}{k}\\ & =\underset{k\to 0}{lim}\frac{f(x+k)g(x)-f(x)g(x+k)}{k\cdot g(x)g(x+k)}\\ & =\underset{k\to 0}{lim}\frac{f(x+k)g(x)-f(x)g(x+k)}{k}\cdot \underset{k\to 0}{lim}\frac{1}{g(x)g(x+k)}\\ & =\underset{k\to 0}{lim}\left[\frac{f(x+k)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+k)}{k}\right]\cdot \frac{1}{[g(x){]}^2}\\ & =\left[\underset{k\to 0}{lim}\frac{f(x+k)g(x)-f(x)g(x)}{k}-\underset{k\to 0}{lim}\frac{f(x)g(x+k)-f(x)g(x)}{k}\right]\cdot \frac{1}{[g(x){]}^2}\\ & =\left[\underset{k\to 0}{lim}\frac{f(x+k)-f(x)}{k}\cdot g(x)-f(x)\cdot \underset{k\to 0}{lim}\frac{g(x+k)-g(x)}{k}\right]\cdot \frac{1}{[g(x){]}^2}\\ & =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{[g(x){]}^2}.\end{array}}$$ The limit evaluation $\underset{k\to 0}{lim}\frac{1}{g(x+k)g(x)}=\frac{1}{[g(x){]}^2}$ is justified by the differentiability of ⁠${\displaystyle g(x)}$⁠, implying continuity, which can be expressed as ⁠${\displaystyle \underset{k\to 0}{lim}g(x+k)=g(x)}$⁠.

Proof using implicit differentiation

Let ⁠${\displaystyle h(x)=\frac{f(x)}{g(x)}}$⁠, so that ⁠${\displaystyle f(x)=g(x)h(x)}$⁠.

The product rule then gives ⁠${\displaystyle f^{\prime }(x)=g^{\prime }(x)h(x)+g(x)h^{\prime }(x)}$⁠.

Solving for ${\displaystyle h^{\prime }(x)}$ and substituting back for ${\displaystyle h(x)}$ gives: $${\displaystyle \begin{array}{cc}{h}^{\prime }(x)& =\frac{f^{\prime }(x)-g^{\prime }(x)h(x)}{g(x)}\\ & =\frac{f^{\prime }(x)-g^{\prime }(x)\cdot \frac{f(x)}{g(x)}}{g(x)}\\ & =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{[g(x){]}^2}.\end{array}}$$

Proof using the reciprocal rule or chain rule

Let ⁠${\displaystyle h(x)=\frac{f(x)}{g(x)}=f(x)\cdot \frac{1}{g(x)}}$⁠.

Then the product rule gives ⁠${\displaystyle h^{\prime }(x)=f^{\prime }(x)\cdot \frac{1}{g(x)}+f(x)\cdot \frac{d}{dx}\left[\frac{1}{g(x)}\right]}$⁠.

To evaluate the derivative in the second term, apply the reciprocal rule, or the power rule along with the chain rule: $${\displaystyle \begin{array}{cc}\frac{d}{dx}\left[\frac{1}{g(x)}\right]& =-\frac{1}{g(x{)}^2}\cdot g^{\prime }(x)\\ & =\frac{-g^{\prime }(x)}{g(x{)}^2}.\end{array}}$$

Substituting the result into the expression gives $${\displaystyle \begin{array}{cc}{h}^{\prime }(x)& =f^{\prime }(x)\cdot \frac{1}{g(x)}+f(x)\cdot \left[\frac{-g^{\prime }(x)}{g(x{)}^2}\right]\\ & =\frac{f^{\prime }(x)}{g(x)}-\frac{f(x)g^{\prime }(x)}{g(x{)}^2}\\ & =\frac{g(x)}{g(x)}\cdot \frac{f^{\prime }(x)}{g(x)}-\frac{f(x)g^{\prime }(x)}{g(x{)}^2}\\ & =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g(x{)}^2}.\end{array}}$$

Proof by logarithmic differentiation

Let ⁠${\displaystyle h(x)=\frac{f(x)}{g(x)}}$⁠. Taking the absolute value and natural logarithm of both sides of the equation gives $${\displaystyle \ln |h(x)|=\ln \left|\frac{f(x)}{g(x)}\right|.}$$

Applying properties of the absolute value and logarithms, $${\displaystyle \ln |h(x)|=\ln |f(x)|-\ln |g(x)|.}$$

Taking the logarithmic derivative of both sides, $${\displaystyle \frac{h^{\prime }(x)}{h(x)}=\frac{f^{\prime }(x)}{f(x)}-\frac{g^{\prime }(x)}{g(x)}.}$$

Solving for ${\displaystyle h^{\prime }(x)}$ and substituting back ${\displaystyle \frac{f(x)}{g(x)}}$ for ${\displaystyle h(x)}$ gives: $${\displaystyle \begin{array}{cc}{h}^{\prime }(x)& =h(x)\left[\frac{f^{\prime }(x)}{f(x)}-\frac{g^{\prime }(x)}{g(x)}\right]\\ & =\frac{f(x)}{g(x)}\left[\frac{f^{\prime }(x)}{f(x)}-\frac{g^{\prime }(x)}{g(x)}\right]\\ & =\frac{f^{\prime }(x)}{g(x)}-\frac{f(x)g^{\prime }(x)}{g(x{)}^2}\\ & =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g(x{)}^2}.\end{array}}$$

Taking the absolute value of the functions is necessary for the logarithmic differentiation of functions that may have negative values, as logarithms are only real-valued for positive arguments. This works because ⁠${\displaystyle \frac{d}{dx}(\ln |u|)=\frac{u^{\prime }}{u}}$⁠, which justifies taking the absolute value of the functions for logarithmic differentiation.

Higher order derivatives

Implicit differentiation can be used to compute the nth derivative of a quotient (partially in terms of its first n−1 derivatives). For example, differentiating ${\displaystyle f=gh}$ twice (resulting in ⁠${\displaystyle f^{″}=g^{″}h+2g^{\prime }{h}^{\prime }+g{h}^{″}}$⁠) and then solving for ${\displaystyle h^{″}}$ yields $${\displaystyle h^{″}={\left(\frac{f}{g}\right)}^{″}=\frac{f^{″}-g^{″}h-2g^{\prime }{h}^{\prime }}{g}.}$$