8.2 Derivatives of Combinations of Functions

]> 8.2 Derivatives of Combinations of Functions

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8.2 Derivatives of Combinations of Functions

If $f = c g$ 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 $f = \frac{g}{h}$ 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 $(c g)' = c g'$ 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 $\frac{g}{h}$ using the previous rule if we know how differentiate $\frac{1}{h}$ , since we have $\frac{g}{h} = g * (\frac{1}{h})$ .

We can find $(\frac{1}{h})'$ by using the fact that $h * (\frac{1}{h}) = 1$ .

By the product rule we obtain $0 = 1' = (h * (\frac{1}{h}))' = h' * (\frac{1}{h}) + h * (\frac{1}{h})'$ .

Rearranging this statement and dividing by $h$ yields $(\frac{1}{h})' = - \frac{h'}{h^{2}}$ .

Exercise 8.1 State the "quotient rule" that follows from these facts, that is the rule for finding $f'$ given $f = \frac{g}{h}$ . Apply it to find $(\frac{\sin x}{\exp x})'$ .

To find $f'$ knowing how to differentiate $g$ and $h$ when $f (x) = g (h (x))$ we need only observe that thinking of $d f$ and $d g$ as differentials we will have $d f = d g$ , while $f'$ means $\frac{d f}{d x}$ , and given $g = g (h)$ knowing $g'$ gives us $\frac{d g}{d h}$ and therefore $\frac{d f}{d h}$ .

To get $\frac{d f}{d x}$ from $\frac{d f}{d h}$ we need to multiply by $\frac{d h}{d x}$ , 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 $\frac{d h}{d x}$ of the tangent line of its graph, after switching the $h$ and $x$ axes we get slope $\frac{d x}{d h}$ .

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 $\frac{1}{h'}$ 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 $\cos x, \tan x, \cot x, \sec x$ and $\csc x$ using their relations to $\sin x$ .

8.3 Use the facts that $x^{1 / n}$ is the inverse function to $x^{n}$ 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.

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 $\frac{1}{h}$ 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 $x^{a}$ 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

(x^{a})' = (\exp (a * \ln (x)))' = \exp (a * \ln (x)) * a (\ln (x))' = (x^{a}) * \frac{a}{x} = a x^{a - 1}

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 $\frac{d (\ln y)}{d y} = \frac{1}{y}$ , 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 $x^{x}$ with respect to $x$ .