<< Chapter < Page | Chapter >> Page > |
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,
□
For $j\left(x\right)=f\left(x\right)g\left(x\right),$ use the product rule to find ${j}^{\prime}(2)$ if $f\left(2\right)=3,{f}^{\prime}\left(2\right)=\mathrm{-4},g\left(2\right)=1,$ and ${g}^{\prime}\left(2\right)=6.$
Since $j\left(x\right)=f\left(x\right)g\left(x\right),{j}^{\prime}\left(x\right)={f}^{\prime}\left(x\right)g\left(x\right)+{g}^{\prime}\left(x\right)f(x),$ and hence
For $j\left(x\right)=({x}^{2}+2)(3{x}^{3}-5x),$ find ${j}^{\prime}(x)$ by applying the product rule. Check the result by first finding the product and then differentiating.
If we set $f\left(x\right)={x}^{2}+2$ and $g\left(x\right)=3{x}^{3}-5x,$ then ${f}^{\prime}\left(x\right)=2x$ and ${g}^{\prime}\left(x\right)=9{x}^{2}-5.$ Thus,
Simplifying, we have
To check, we see that $j\left(x\right)=3{x}^{5}+{x}^{3}-10x$ and, consequently, ${j}^{\prime}\left(x\right)=15{x}^{4}+3{x}^{2}-10.$
Use the product rule to obtain the derivative of $j\left(x\right)=2{x}^{5}\left(4{x}^{2}+x\right).$
${j}^{\prime}\left(x\right)=10{x}^{4}\left(4{x}^{2}+x\right)+\left(8x+1\right)\left(2{x}^{5}\right)=56{x}^{6}+12{x}^{5}.$
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
Let $f(x)$ and $g(x)$ be differentiable functions. Then
That is,
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.
Use the quotient rule to find the derivative of $k\left(x\right)=\frac{5{x}^{2}}{4x+3}.$
Let $f\left(x\right)=5{x}^{2}$ and $g\left(x\right)=4x+3.$ Thus, ${f}^{\prime}\left(x\right)=10x$ and ${g}^{\prime}\left(x\right)=4.$ Substituting into the quotient rule, we have
Simplifying, we obtain
Find the derivative of $h\left(x\right)=\frac{3x+1}{4x-3}.$
${k}^{\prime}\left(x\right)=-\frac{13}{{(4x-3)}^{2}}.$
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.
If $k$ is a negative integer, then
If $k$ is a negative integer, we may set $n=\text{\u2212}k,$ so that n is a positive integer with $k=\text{\u2212}n.$ Since for each positive integer $n,{x}^{\text{\u2212}n}=\frac{1}{{x}^{n}},$ 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,
Notification Switch
Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?