The Product rule and quotient rule

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\CancelColor }[0]{} \newcommand {\cancelto }[0]{1} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$ $% This file was *autogenerated* from digInProductRuleAndQuotientRule.sagetex.sage with % sagetex.py version 2015/08/26 v3.0-92d9f7a %1523609e348dcad0fae096f152f78176% md5sum of corresponding .sage file (minus "goboom","current_tex_line",and pause/unpause lines)$

Here we compute derivatives of products and quotients of functions

The product rule

Consider the product of two simple functions, say $f(x)\cdot g(x)$ where $f(x)=x^2+1$ and $g(x)=x^3-3x$ . An obvious guess for the derivative of $f(x)g(x)$ is the product of the derivatives:

$\begin{align*} f'(x)g'(x) &= (2x)(3x^2-3)\\ &= 6x^3-6x. \end{align*}$

Is this guess correct? We can check by rewriting $f$ and $g$ and doing the calculation in a way that is known to work. Write with me

$\begin{align*} f(x)g(x) &= (x^2+1)(x^3-3x)\\ &=x^5-3x^3+x^3-3x\\ &=x^5-2x^3-3x. \end{align*}$

Hence $\ddx f(x) g(x) = \ddx (x^5 - 2x^3 - 3x) = 5x^4-6x^2-3,$ so we see that $\ddx f(x) g(x) \ne f'(x)g'(x).$ So the derivative of $f(x)g(x)$ is not as simple as $f'(x)g'(x)$ . Never fear, we have a rule for exactly this situation.

The product rule If $f$ and $g$ are differentiable, then $\ddx f(x)g(x) = f(x)g'(x)+f'(x)g(x).$

Let’s return to the example with which we started.

Let $f(x)=(x^2+1)$ and $g(x)=(x^3-3x)$ . Compute: $\ddx f(x)g(x)$

Write with me $\begin{align*} \ddx f(x)g(x) &= f(x)g'(x) + f'(x)g(x)\\ &=(x^2+1)(\answer[given]{3x^2-3}) + (\answer[given]{2x})(x^3-3x). \end{align*}$

We could stop here, but we should show that expanding this out recovers our previous result. Write with me

$\begin{align*} (x^2+1)&(3x^2-3) + 2x(x^3-3x)\\ &= 3x^4-3x^2 +3x^2 -3 + 2x^4-6x^2\\ &=\answer[given]{5x^4-6x^2-3}, \end{align*}$

which is precisely what we obtained before.

Now that we are pros, let’s try one more example.

The quotient rule

We’d like to have a formula to compute $\ddx \frac {f(x)}{g(x)}$ if we already know $f'(x)$ and $g'(x)$ . Instead of attacking this problem head-on, let’s notice that we’ve already done part of the problem: $f(x)/g(x)= f(x)\cdot (1/g(x))$ , that is, this is really a product, and we can compute the derivative if we know $f'(x)$ and $(1/g(x))'$ . This brings us to our next derivative rule.

The quotient rule If $f$ and $g$ are differentiable, then $\ddx \frac {f(x)}{g(x)} = \frac {f'(x)g(x)-f(x)g'(x)}{g(x)^2}.$

Compute: $\ddx \frac {x^2+1}{x^3-3x}$

Write with me $\begin{align*} \ddx \frac{x^2+1}{x^3-3x} &= \frac{2x(\answer[given]{x^3-3x})-(\answer[given]{x^2+1})(3x^2-3)}{(\answer[given]{x^3-3x})^2}\\ &=\frac{-x^4-6x^2+3}{(\answer[given]{x^3-3x})^2}. \end{align*}$

It is often possible to calculate derivatives in more than one way, as we have already seen. Since every quotient can be written as a product, it is always possible to use the product rule to compute the derivative, though it is not always simpler.

Compute: $\ddx \frac {625-x^2}{\sqrt {x}}$ in two ways. First using the quotient rule and then using the product rule.

First, we’ll compute the derivative using the quotient rule. Write with me $\ddx \frac {625-x^2}{\sqrt {x}} = \frac {\left (-2x\right )\left (\answer [given]{\sqrt {x}}\right ) - (\answer [given]{625-x^2})\left (\frac {1}{2}x^{-1/2}\right )}{\answer [given]{x}}.$

Second, we’ll compute the derivative using the product rule:

$\begin{align*} \ddx \frac{625-x^2}{\sqrt{x}} &= \ddx \left(625-x^2\right)x^{-1/2}\\ =\left(625-x^2\right)&\left(\answer[given]{\frac{-x^{-3/2}}{2}}\right)+ (\answer[given]{-2x})\left(x^{-1/2}\right). \end{align*}$

With a bit of algebra, both of these simplify to $-\frac {3x^2+625}{2x^{3/2}}.$