Some teachers have a template for students to complete for the first few assignments and then revert to using the template when student accuracy slides. If you use the power rule plus the product rule, you often must find a common denominator to simplify the result. Forgetting to square the denominator is also common. There are two reasons why the quotient rule can be superior to the power rule plus product rule in differentiating a quotient: It preserves common denominators when simplifying the result. Adding terms in the numerator is a common mistake. Perfect practice makes perfect, so expect and require correct work on these problems. 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. Non-calculator derivatives are typically less complex than those that require numerical derivatives.īeware! Students commonly reverse the ‘order of operations’ for this derivative and first multiply the numerator by the derivative of the denominator. Having developed and practiced the product rule, we now consider differentiating quotients of functions. The Quotient Rule is present throughout all sections of the exam. The best strategy for any student is the one that works best for them. If you have found success with a particular mnemonic or method, continue using that strategy. The internet is full of tricks and gimmicks and rhymes for remembering the Quotient Rule (low-de-high high-de-low, or “Bottoms up!”). This topic presents a great opportunity to continue a discussion about the domain of the denominator. To impress upon students the complexity of the Quotient Rule, today we will let them make a conjecture, test it on their calculators, and then realize they are wrong: we are going to “break” the tool they try to use, thus creating the need for the formula. Right next to the chain rule, this is one of the most commonly mis-applied derivative rules out there. They will encounter many different functions in the numerators and denominators of “quotient functions,” so remembering the process for dealing with those functions is really important. ![]() While the derivatives of sin x, or cos x, or ln x are memorized, the Quotient Rule provides students with a process (as did the Product Rule). Simplifying the numerator, you get -3x plus 6 over root x times x plus 2 quantity squared.We meet the last “procedural” derivative rule of Unit 2 today. I think this looks nicer than the one we had before. You do get a root x in the denominator, but that's okay. You'll have 3 times x plus 2, so that's 3x plus 6 minus. So multiplying the root x through the numerator, you get a cancellation with this guy. Simplifying complex fractions is something that many students are required to do. Actually one thing that you can do about that to get rid of fractions within fractions, these are called complex fractions, is you can multiply the top and bottom of this answer with root x. So this is a little complicated, because we've got a tiny fraction here. So 6 root x times the derivative of x plus 2 which is 1, over the square of what's below. But let me just state the quotient rule right now. So the derivative of 6 root x is going to be 6 times this. The quotient rule, I'm gonna state it right now, it could be useful to know it, but in case you ever forget it, you can derive it pretty quickly from the product rule, and if you know it, the chain rule combined, you can get the quotient rule pretty quick. Low d high, x plus 2 times the derivative of this function. X to the -½ is the same as 1 over the square root of x. You get that by thinking of the square root of x as x to the ½. Let me just write this on the side, the derivative. Remember the derivative of the square root of x is 1 over 2 root x. Here is my low function, here is my high function. So I'm going to use the quotient rule on this. My final answer is 2e to the x over e to the x plus 1 quantity squared. Then minus, minus gives me +1 times e to the x. Then I have e to the x times e to the x which is another e to the 2x, so I can see that those are going to cancel. This becomes, I have to the x times e to the x, which is e to the x quantity squared, or e to the 2x plus e to the x minus. So e to the x, now I'm remembering plus in the denominator, plus 1. So e to the x minus 1 times the derivative of this guy which is also e to the x over, the square of what's below. It's going to be low d high e to the x times the derivative of e to the x minus 1, which is e to the x, minus high d low. ![]() The derivative, dy/dx, is going to be, and I have to use the quotient rule for this guy. So the numerator and the denominator are very similar, but this is still an interesting function to look at if you graph it. Here we're asked to differentiate y equals e to the x minus 1 over e to the x plus 1.
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