what is the relation of f ' to g ' and h '?|
|
||
If f = cg what is the relation between f ' and g ', where c is a constant?
If f = g + h or f = g - h what is the relation of f ' to g' and h '?
If f = g * h same question?
If
what is the relation of f ' to g ' and h '?
If f(x) = g(h(x)) same question.
If f is the inverse function to g, f = g-1?
If f obeys the equation g(f(x)) = 0, in any open interval containing x, how can we find f '(x) in terms of g and g '?
We know from the definition of derivate that (cg) ' = cg ' when c is a constant.
We also know by the fundamental principle described in Chapter 6, that contributions to derivatives from different sources merely add.
We can immediately deduce how to differentiate sums differences and products.
Rule for differentiating sums: (g + h)' = g' + h'. Also (g - h)' = g' - h'.
Rule for differentiating products: (g * h)' = g * h' + g' * h.
We can obtain the rule for finding the derivative of 
using the previous rule if we know how differentiate 
,
since we have 
We can find 
by using the fact that 
By the product rule we obtain 
Rearranging this statement and dividing by h yields 
Exercise 8.1 State the "quotient rule" that follows from these
facts, that is the rule for finding f ' given .
Apply it to find 
To find f ' knowing how to differentiate g and h when f(x) = g(h(x)) we need
only observe that thinking of df and dg as differentials we will have df = dg,
while f ' means 
,
and given g = g(h) knowing g' gives us 
and therefore 
.
To get
from
we need to multiply by 
,
so we get "the Chain rule" (g(h(x))' = g'(h) * h'(x) where g' (h) is evaluated at h = h(x).
To find the derivative of the inverse function to h(x), you need only to observe
that the inverse function is obtained by switching x and y axes; since the derivative
of h is the slope
of the tangent line of its graph, after switching the h and x axes we get slope
Thus the derivative of the inverse function (h-1(x)) to h at argument h(x) is the reciprocal of the derivative of h with respect to x and argument x.
We get (h-1(x))' evaluated at x = h(z) is 
with h' evaluated at z. This sounds worse than it is.
Another way to get the same result is to apply the chain rule to the alternate
definition of the inverse function: h-1(h(x)) = x.
By the chain rule we get 1 = x' = (h-1(h(x))' = (h-1)'
* h'(x), where h-1 is evaluated at h(x); again the conclusion is
that (h-1)' evaluated at h(x) is the reciprocal of h'(x).
Exercises:
8.2 Find the derivatives of cos x, tan x, cot x, sec x and csc x using their relations to sin x.
8.3 Use the facts that x1/n is the inverse function to xn to find (x1/n)' and also find the derivatives of the inverse functions to exp(x), sin x, and tan x (namely ln(x), arcsin x, and arctan x) by applying the "inverse rule" just described.
Now suppose g(f(x)) = 0 over an interval containing x.
We can then apply the chain rule to find (g(f(x)))' = 0' = 0 = g'(f) * f
'(x), and this equation will determine f ' in terms of f. This is actually
the general
idea used above to evaluate the derivatives both of
and h-1 (the reciprocal and the inverse functions to h).
Is this all we need to differentiate? The answer is yes.
Notice that there are really only two rules that we have invoked here which allow us to differentiate all standard functions.
One is the multiple occurrence rule, which allows us to treat separate occurrences of a variable separately and add their individual derivatives up to get the whole derivative.
The second is the chain rule, which notes that the derivative is a slope which is a ratio of changes, so that changing the independent variable, which changes the denominator of the slope, requires a change in the derivative of the ratio of the old denominator to the new one.
To illustrate how these rules can be employed we deduce the "power rule" from them.
Recall that we can define the function xa for any a to be exp(a * ln(x)).
We can use the fact that exp(x) is its own derivative, with the chain rule to tell us
OK, how did we get the formula for the derivative of ln(x)?
Well, ln(x) is the inverse function to exp(x). This means that if y = exp(x) then x = ln(y). But exp(x) is its own derivative, which means y' is y when y = exp(x).
Since the derivative of an inverse function is the reciprocal of the derivative
of the original function, we get 
,
which is the formula we used.
There is one other way that variables occur that we have not yet encountered but should be mentioned. They can and often do occur in upper (or lower) limits of integration. You will learn how to differentiate a function of this kind when that concept is discussed. You should be aware here though, that when a variable occurs both as such a limit and as an ordinary function, you can invoke the "separate occurrence" rule to treat these occurrences each individually and add the contributions to the derivative coming from each to get the entire derivative.
Exercise 8.4 Find the derivative of xx with respect to x.
|
|