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In a perfect world, everything would be easy and there would be tricks to calculus. However, this is not a perfect world. So, teachers do not always teach the easiest ways to do things. There are several reasons for this.

1. There just isn't enough time.

2. Some tricks, like the ones on this page, teach you to get answers without learning the technique that the instructor is required to teach you and test you on.

3. Most tricks are useful only for very specific types of problems. You will often find that learning tricks takes more time and energy to learn than they are worth. Therefore the time and energy to learn them is better spent on learning the actual technique required to pass the course.

As an instructor, I believe that each of these are valid reasons not to teach these tricks. In my opinion, tricks keep you from learning the basics and can make future use of calculus difficult when you run into a problem that does not fit a specific trick.

That said, if you want to take the time to learn tricks, you can use them to check your answers. This is a very good use of tricks and, as an instructor, I encourage you to do this. However, do not expect to get full credit on your assignments and exams if you use these techniques as your answers, since your instructor probably requires you to use another specific technique. Check with your instructor to see what they require.

Let's start with some Quotient Rule tricks. The Quotient Rule can appear to be quite complicated at first, which is why you will see a lot of tricks trying to get around using this rule. However, if you spend a few extra minutes actually learning the Quotient Rule, many of these tricks will not be required.

Although I wouldn't really call this a trick, it is a handy shortcut when you are presented with a rational expression with a constant in the numerator and you are asked to use the quotient rule. He calls this the Reciprocal Rule. It looks like this. \[ \frac{d}{dx} \left[ \frac{1}{f(x)} \right] = \frac{-f'(x)}{[f(x)]^2} \]

Here is a video showing several examples and expanding the idea to apply to rational terms that have a constant, not just the number one, in the numerator.

Okay, here is why this works.

For this one, you have a rational function and the Quotient Rule would apply. However, for very specific functions, this one might save you some time. Be careful with this one. It is easy to get tripped up. Here it is. \[ \frac{d}{dx}\left[ \frac{af(x)+b}{cf(x)+d} \right] = \frac{f'(x)(ad-bc)}{[cf(x)+d]^2} \] In this formula, \(a\), \(b\), \(c\) and \(d\) are all constants and the function \(f(x)\) in the numerator and denominator must be the same. You can see that this trick is quite specialized. But keep in mind that your instructor will probably not give you full credit for using this as your answer.

Here is why this works.

I wouldn't exactly call this a trick. It's more like something you pick up as you work practice problems. Here it is. \[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \] I like how he shows in this video how to extend this formula when the chain rule is required.

Here is why this works.

Here is a trick he uses to bypass implicit differentiation for partial derivatives. You will not see partial derivatives until calculus 3. Here is the formula. \[ \frac{d}{dx}f(x,y) = \frac{-f_x}{f_y} \]

If you watched that entire video clip, he says at the end that you will need to use your regular procedure anyway to get full points. So I would say that this formula could be used only to check your answer.