tom.h.wilson Dept. Geology and Geography West Virginia University
Functions of the type Recall our earlier discussions of the porosity depth relationship
But it worked out a little differently for Don’t get the derivatives of these exponential functions confused with the power rule
The book works through the differentiation of y = x 2, so let’s try y =x 4. multiplying that out -- you get...
Remember this idea of dy and dx is that the differential changes are infinitesimal - very small. So if dx is (that’s 1x10 -4 ) then (dx) 2 = (or 1x10 -8 ) (dx) 3 = 1x and (dx) 4 = 1x So even though dx is infinitesimally small, (dx) 2 is orders of magnitude smaller
so that we can just ignore all those terms with (dx) n where n is greater than 1. Our equation gets simple fast Also, since y =x 4, we have and then -
Divide both sides of this equation by dx to get This illustrates the general form of the power rule –
is Again, as a reminder, remember that the constant factors in an expression carry through the differentiation. This is obvious when we consider the derivative - which - in general for
Examining the effects of differential increments in y and x we get the following
Don’t let negative exponents fool you either. If n is -1, for example, we still have And the result is In this case
Given the function - what is? We just differentiate f and g individually and take their sum, so that
Take the simple example - what is? What are the individual derivatives of and?
let then - We just apply the power rule and obtain We know from the forgoing note that the c disappears.
We use the power rule again to evaluate the second term, letting g = (ax 4 +b) Thus -
Differences are treated just like sums so that is just
Recall how to handle derivatives of products ? or
Removing explicit reference to the independent variable x, we have Going back to first principles, we have Evaluating this yields Since df x dg is very small and since y=fg, the above becomes -
Which is a general statement of the rule used to evaluate the derivative of a product of functions The quotient rule is just a variant of the product rule, which is used to differentiate functions like
The quotient rule states that And in most texts the proof of this relationship is a rather tedious one. The quotient rule is easily demonstrated however, by rewriting the quotient as a product and applying the product rule. Consider
We could let h=g -1 and then rewrite y as Its derivative using the product rule is just and substitution yields It’s actually necessary to incorporate the chain rule into this since
Multiply the first term in the sum by g/g (i.e. 1) to get > Which reduces to i.e. the quotient rule
The derivative of an exponential function Given > In general for If express a as e n so that then Note
Sinceand in general a can be thought of as a general base. It could be 10 or 2, etc.
The derivative of logarithmic functions Given > We’ll talk more about these special cases after we talk about the chain rule.
Differentiating functions of functions - Given a functionwe consider writecompute Then computeand take the product of the two, yielding
We can also think of the application of the chain rule especially when powers are involved as working form the outside to inside of a function
Where Derivative of the quantity squared viewed from the outside. Again use power rule to differentiate the inside term(s)
Using a trig function such as let then Which reduces toor just
In general if then
Returning to those exponential and natural log cases - we already implemented the chain rule when differentiating h in this case would be ax and, from the chain rule, becomesor and finally since and
For functions like we follow the same procedure. Letand then From the chain rule we have hence
Thus for that porosity depth relationship we were working with -
For logarithmic functions like We combine two rules, the special rule for natural logs and the chain rule. Let Chain rule Log rule then and so
For next Tuesday answer question 8.8 in Waltham (see page 148). Find the derivatives of