Tom.h.wilson Dept. Geology and Geography West Virginia University.

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Presentation transcript:

tom.h.wilson Dept. Geology and Geography West Virginia University

Estimating the rate of change of functions with variable slope

The book works through the differentiation of y = x 2, so let’s try y =x 4. multiplying that out -- you get...

Remember the idea of the dy and dx is that they represent differential changes that 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 very 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 is just another illustration of what you already know as the power rule,

is Just as a footnote, 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. If n is -1, for example, we still have or just

Given the function - what is? We just differentiate f and g individually and take their sum, so that

Take the simple example - what is? We can rewrite Then just think of the derivative operator as being a distributive operator that acts on each term in the sum.

Where then - On the first term apply the power rule What happens to and ?

Successive differentiations yield Thus -

Differences are treated just like sums so that is just

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 (the angle is another function 2ax)

In general if then

How do you handle derivatives of functions like ? or The products and quotients of other functions

Removing explicit reference to the independent variable x, we have Going back to first principles, we have Evaluating this yields Since dfdg is very small we let it equal zero; 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 The proof of this relationship can be tedious, but I think you can get it much easier using the power rule Rewrite the quotient as a product and apply the product rule to y as shown below

We could let h=g -1 and then rewrite y as Its derivative using the product rule is just dh = -g -2 dg and substitution yields

Multiply the first term in the sum by g/g (i.e. 1) to get > Which reduces to the quotient rule

Functions of the type Recall our earlier discussions of the porosity depth relationship

Refer to comments on the computer lab exercise. Derivative concepts

Between 1 and 2 kilometers the gradient is km -1

As we converge toward 1km,  /  z decreases to km -1 between 1 and 1.1 km depths.

What is the gradient at 1km? What is ?

This is an application of the rule for differentiating exponents and the chain rule

Next time we’ll continue with exponentials and logs, but also have a look at question 8.8 in Waltham (see page 148). Find the derivatives of

tom.h.wilson Dept. Geology and Geography West Virginia University

Differentiating exponential and log functions

Returning to those exponential and natural log cases – recall we implement 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 -

The derivative of logarithmic functions Given > We’ll talk more about these special cases after we talk about the chain rule.

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

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.