3.3 Differentiation rules  (Page 2/14)

By adding and subtracting $f\left(x\right)g(x+h)$ in the numerator, we have

After breaking apart this quotient and applying the sum law for limits, the derivative becomes

Rearranging, we obtain

By using the continuity of $g\left(x\right),$ the definition of the derivatives of $f(x)$ and $g(x),$ and applying the limit laws, we arrive at the product rule,

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The quotient rule

Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that

The proof of the quotient rule    is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form $x^k$ where $k$ is a negative integer.

Proof

If $k$ is a negative integer, we may set $n=\text{−}k,$ so that n is a positive integer with $k=\text{−}n.$ Since for each positive integer $n,x^{\text{−}n}=\frac{1}{x^n},$ we may now apply the quotient rule by setting $f\left(x\right)=1$ and $g\left(x\right)=x^n.$ In this case, $f^{\prime }\left(x\right)=0$ and $g^{\prime }\left(x\right)=n{x}^{n-1}.$ Thus,