» Required Reading » Table of Contents » Chapter 8 8.2 Derivatives of Combinations of Functions
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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 f = g / h what is the relation of f ' to g ' and h '?
If f(x) = g(h(x)) …
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 definition that (cg) ' = cg '. 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 g / h using the previous rule if we know how differentiate 1 / h, since we have g / h = g * (1 / h).
We can find (1 / h)' by using the fact that h * (1 / h) = 1
By the product rule we obtain 0 = 1' = (h * (1 / h))' = h' * (1 / h) + h * (1 / h)'
Rearranging this statement and dividing by h yields (1/h)' = - h' / h².

Exercise 8.1: State the “quotient rule” that follows from these facts, that is the rule for finding f ' given f = g / h. Apply it to find ((sin x) / (exp x))'.

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 df / dx, and given g = g(h) knowing g' gives us dg / dh and therefore df / dh. To get df / dx we need multiply df / dh by dh / dx, 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 observe that the inverse function is obtained by switching x and y axes; since the derivative of h is the slope dh / dx of the tangent line of its graph, after switching the h and x axes we get slope dx / dh. 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 1 / h' 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, of tan x, cot x, sec x and csc x using their relations to sin x.

8.3 Use the facts that x^1/n is the inverse function to xⁿ to find (x^1/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.

And if 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 1 / h and h^-1 (the reciprocal and the inverse functions to h)
Is this all we need to differentiate? The answer is yes.