Today we’ll look at some tips on how to figure out derivatives in calculus. Disclaimer: I’m no math pro, so this might not all be accurate.

*Update*

A really nice basic demonstration of how the to use the difference quotient and how it is related to some of the basic rules of differentiation is located at http://www.intmath.com/Differentiation/3_Derivative-first-principles.php.

One important thing is to pin down what the independent variable(s) is/are in the equations you will be differentiating–generally denoted in letters at the end of the alphabet, like x, y, or z in many situations, such if you’re given the functions f(x), g(y), or h(z)–and figuring out what variables you’ll be differentiating with respect to. This is important, because for any constant in the function (generally* *denoted by letters at the beginning of the alphabet such as a, b, c, d, etc in many situations) the derivative will be zero (unless the constant is multiplied by the variable you’re differentiating with respect to)–examine the constant rule:

*Constant rule*: if*f*is the constant function*f*(*x*) =*c*, for any number*c*, then for all*x**.*

and the constant factor/constant multiple rules of differentiation.

When looking at functions with more than one variable, the variables you’re not differentiating with respect to are also treated like constants, like when taking partial derivatives.

Here are some examples of where pinning down the constants, and realizing that their derivative will be zero, will be useful, in combination with a handful of differentiation rules.

For example, in this explanation of the proof of the second mean value theorem in the book Mathematics for Economists, using Rolle’s theorem, part of the proof involves taking the second derivative of the equation , where a is a constant.

Now, taking the derivative of this equation involves the sum rule, also called the combination rule, the product rule, the power rule, the constant rule, and the constant factor/constant multiple rules of differentiation, as well as the basic distributive property of multiplication.

So, to get R'(x), first of all, you can separately take the derivatives of all the terms separated by plus (+) signs, according to the sum rule/the combination rule. So, taking the derivatives of each of the terms:

First, d/dx(f(x))=f'(x). Next, d/dx(f(a))=0 since a is a constant, by the constant rule–since a is a constant, f(a) also is a constant, since for each input of a function there is a unique output, and the constant rule says that the derivative of a constant is 0. Next, we have (x-a)f'(a). Using the distributive property of multiplication, this factors into xf'(a) + -af'(a). Next, to find d/dx(xf'(a)), we use the constant factor/constant multiple rule. Since a is a constant, on its own f'(a) would be zero, but since it’s multiplied by x, we multiply it by the derivative of x which is 1, so d/dx(xf'(a))=f'(a). Next, we have d/dx(-af'(a)). Here both a and f'(a) are constants, so both d/dx(a) and d/dx(f'(a)) are 0, and d/dx(-af'(a))=0. Next we have . First we can multiply out , from which we get , so we get (Note: we could also have used the composite function/chain rule here instead of multiplying out , see the end of the post for more details). Now taking the derivative of this, d/dx ( )=-2Kx by the power rule and the constant factor/constant multiple rule, and d/dx(2Kax)=2ka by the same rules, and d/dx()=0 since a is a constant, by the constant rule.

This leaves us with R'(x)=f'(x) + 0 + -f'(a) + 0 + -2Kx + 2Ka. Now, to take the derivative of R'(x), which will be the second derivative of R(x), which will be R”(x).

First, d/dx(f'(x))=f”(x). Next, d/dx(-f'(a))=0 since a is a constant. Next, d/dx(-2Kx)=-2K by the power rule and the constant factor/constant multiple rule. Next, d/dx(2Ka)=0 since 2, K, and a are constants, and the derivative of a constant equals 0.

This leaves us with R”(x)=f”(x) + -2K. Further on in the proof, using Rolle’s theorem, you find that there is a number d such that R”(d)=0. Thus, R”(d)=f”(d) + -2K, which equals 0=f”(d) +-2K, which leaves us with f”(d)=2K. The rest of the proof is in the book.

Hopefully this is all correct (it might not be, I’m no math pro), and demonstrates how all sorts of different differentiation rules are used in calculus problems!

- Note 1: we could have used the composite function/chain rule instead of multiplying out (see the section on composition of functions also). By the composite function rule, for the derivative of we would set q(x)=u=(x-a) , and p(u)=y=. We would set f(x)=p(q(x))=y. Then dy/dx=du/dx * dy/du. Now, du/dx, where u=(x-a), equals 1 by the sum rule/the combination rule, the power rule, and the constant rule. Next, dy/du where y= is 2u, by the power rule. Then we want dy/dx=du/dx * dy/du. So that’s 2u * 1. Substituting in u=(x-a) we get 2(x-a) or 2x-2a. So to find the derivative of , we use the constant factor/constant multiple rule for the -K, and the derivative of , is -2K(x-a) or -2Kx+2Ka like we figured out above.

- Note 2: We didn’t even get to the product rule/multiplication rule!!! The product rule can be used in equations where you have more than one term containing the independent variable that you are differentiating with respect to–you can use the product rule when those terms are multiplied together. You can read more about the product rule in the Mathematics for Economists book here. For example, if you had , you could easily multiply the terms together to get then use the power rule to get , but for the sake of example let’s use the product rule. If we set x=u and , and f(x)=uv, using product rule, f'(x)=v * du/dx + u * dv/dx. Here, du/dx=1 by the power rule, and dv/dx=2x by the power rule. Then , which shows you that that’s what we would have gotten if we had just multiplied the terms together and taken the derivative using the power rule. The product rule can be helpful in cases where it would be simpler to product rule than it would be to multiply the terms together first then take the derivative.

The Rushmore math picture is from this Math at the Movies webpage.

Also see:

- Understanding Integration, Integrals, Antiderivatives, and their Relationship to Derivatives: the Fundamental Theorem of Calculus
- Learning About Critical Points, Points of Inflection, First Derivatives, the Second Derivative Test, Convacity, Graphing Polynomials, etc.

Filed under: math, Uncategorized | Tagged: calculus, derivatives, differentiation, math |

keith beard, on January 4, 2010 at 1:13 am said:its all totally greek to me.