With the constant multiple and sum rules, the product and quotient rules allow us to compute the derivative of each function consisting of sums, constant multiples, products, and quotients of basis functions. For example, if (F) is the form If a function is a sum, product, or quotient of simpler functions, then we can use sum, product, or quotient rules to distinguish them in terms of simpler functions and their derivatives. There is a lot of emphasis on the difference quotient since it`s the AP test and all that, but honestly, using the power rule plus the product rule is so much easier and gives you the same answer and slope. What am I missing here? For example, if you are working on differential equations, it is often useful to recognize an expression of the form $u v` – u` v$ with respect to the derivation of a quotient. Sometimes you need to calculate the derivative of a quotient with a constant numerator, such as $ds 10/x^2$. Sure, you can use the quotient rule, but it`s usually not the easiest method. If we use it here, we get $${dover dx}{10over x^2}={x^2cdot 0-10cdot 2xover x^4}= {-20over x^3},$$, since the derivative of 10 is 0. But it`s easier to do: $${dover dx}{10over x^2}={dover dx}10x^{-2}=-20x^{-3}.$$ Admittedly, $ds x^2$ is a particularly simple denominator, but we will see that a similar calculation is usually possible. Another approach is to remember that $${dover dx}{1over g(x)}={-g`(x)over g(x)^2},$$ but this requires additional memorization. With this formula, $${dover dx}{10over x^2}=10{-2xover x^4}.$$ Note that we first use the linearity of the derivative to drag the 10 forward. It maintains common denominators when simplifying the result. If you use the power rule plus the product rule, you often need to find a common denominator to simplify the result.
The trade-off here is between more calculus/less algebra and less calculus/more algebra. As I find calculus less prone to errors than algebra, I would opt for the more calculus/minus algebra approach. Quotient rule: $${dover dx}{sqrt{625-x^2}oversqrt{x}} = {sqrt{x}(-x/sqrt{625-x^2})-sqrt{625-x^2}cdot 1/(2sqrt{x})over x}.$$ Note that we used $ds sqrt{x}=x^{1/2}$ to calculate the derivative of $ds sqrt{x}$ by the power rule. What is the derivative of $ds (x^2+1)/(x^3-3x)$? In general, we would like to have a formula to calculate the derivative of $f(x)/g(x)$ if we already know $f`(x)$ and $g`(x)$. Instead of tackling this problem head-on, we should note that we have already solved part of the problem: $f(x)/g(x)= f(x)cdot(1/g(x))$, i.e. it is “really” a product, and we can calculate the derivative if we know $f`(x)$ and $(1/g(x))`$. The only new information we need is $(1/g(x))`$ in terms of $g`(x)$. As for the product rule, let`s configure this and see how far we can go: $$ eqalign{ {dover dx}{1over g(x)}&=lim_{Delta xto0} {{1over g(x+Delta x)}-{1over g(x)}overDelta x}cr &=lim_{Delta xto0} {{g(x)-g(x+Delta x)over g(x+Delta x)g(x)}overDelta x}cr &=lim_{Delta xto0} {g(x)-g(x+Delta x)over g( x+Delta x)g(x)Delta x}cr &= lim_{Delta xto0} -{g(x+Delta x)-g(x)over Delta x} {1over g(x+Delta x)g(x)}cr &=-{g`(x)over g(x)^2}cr }$$ Now we can merge this with the product rule: $${dover dx}{f(x)over g(x)}=f(x){-g`(x)over g(x)^2}+f`(x){1over g(x)}={-f(x)g`(x)+f`(x) g(x) on g(x)^2}= {f`(x)g(x)-f(x)g`(x)over g(x)^2}. $$ Example 3.4.2 Find the derivative of $ds sqrt{625-x^2}/sqrt{x}$ in two ways: using the quotient rule and using the product rule. Product rule: $${dover dx}sqrt{625-x^2} x^{-1/2} = sqrt{625-x^2} {-1over 2}x^{-3/2}+{-xover sqrt{625-x^2}}x^{-1/2}. There are two reasons why the quotient rule may be superior to the power rule plus the product rule when it comes to distinguishing a quotient: I am confused by the question.
In AP Calc classes, you need to know both, but you can usually use the free answer to derive equations using the method of your choice. You may need to be able to recognize a difference quotient in a multiple-choice or free-response method, and it`s important to understand the intuition behind it! However, you certainly don`t have to solve all the problems. Hope this helps! There are some, very few, integrals that succumb to the quotient rule (vice versa). If you are very familiar with the quotient rule, you can sometimes recognize these integrals. If you`re having trouble remembering the quotient rule, here`s a mnemonic device that can help: “low, dee-high minus high, dee-bas above the square of what`s below.” There are countless variations of this mnemonic device, but I think the quotient rule is worth staying in your arsenal, although it`s technically useless. You should have it for the same reason that you should memorize the square formula, although if you know how to complete the square, you do not need the square formula: it is faster and less prone to errors. then (F) is a quotient in which the numerator is a sum of constant multiples and the denominator is a product. This derivative of (F) can be found by applying the quotient rule and then using the sum and constant multiple rules to distinguish the numerator and the product rule to differentiate the denominator. It is often possible to calculate derivatives in more than one way, as we have already seen. Since any quotient can be written as a product, it is always possible to use the product rule to calculate the derivative, although this is not always easier.
With a bit of algebra, both simplify to $$-{x^2+625over 2sqrt{625-x^2}x^{3/2}}.$$ For example, if you want to distinguish $$f(x)=frac{x^2+4x-2}{x-1},$$, you can use $$f`(x)=frac{d}{dx}[(x^2+4x-2)(x-1)^{-1}]=(2x+4)(x-1)^{-1}-(x^2+4x-2)(x-1)^{-2}, $$ or you can make $$ f`(x)=frac{(x-1)(2x+4)-(x^2+4x-2)(1)}{(x-1)^2}.