Derivatives

Section 1.2 Derivatives

In this section, we study various operations on differentiable functions that preserve differentiability. The derivative of a function obtained by these operations can be expressed in terms of its constituent functions and their derivatives. These expressions are the so-called "rules of differentiation".

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Subsection 1.2.1 Rules of Differentiation

Proposition 1.6.

Suppose

f (x)

and

g (x)

are differentiable at

x = x_{0}

then so are their sum and product. Moreover,

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\begin{aligned} (f + g)^{'} (x_{0}) & = f^{'} (x_{0}) + g^{'} (x_{0}) \\ (f g)^{'} (x_{0}) & = f^{'} (x_{0}) g (x_{0}) + f (x_{0}) g^{'} (x_{0}) \end{aligned}

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Viewing these derivatives as functions, we write

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\begin{aligned} (sum rule) & (f + g)^{'} & = f^{'} + g^{'} \\ (product rule) & (f g)^{'} & = f^{'} g + f g^{'} \end{aligned}

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Since the derivative of a constant

c

is zero, in particular

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\begin{aligned} (f + c)^{'} & = f^{'} \\ (c f)^{'} & = c f^{'} + c^{'} f = c f^{'} \end{aligned}

Also since

f - g = f + (- 1) g,

(f - g)^{'} = f^{'} + ((- 1) g)^{'} = f^{'} - g^{'} .

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Example 1.7.

Applying the product rule to

f (x) = g (x) = x,

we conclude that

(f g) (x) = x^{2}

is differentiable. Moreover,

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(x^{2})^{'} = x^{'} (x) + x (x^{'}) = 2 x .

That is exactly what we have found using the definition.

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More generally, one can check by induction that

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\begin{matrix} (x^{n})^{'} = n x^{n - 1} \end{matrix}

for every positive integer

n .

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In fact, not only the non-negative powers of

x

but for any real number

r,

the function

x \mapsto x^{r}

is differentiable and that

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\begin{matrix} (power rule) & (x^{r})^{'} = r x^{r - 1} \end{matrix}

We will be justifying this later.

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Since polynomial functions can be obtained from constants and the identity function by taking sum and product, they are differentiable. Moreover,

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\begin{matrix} (1.3) & {(\sum_{k = 0}^{n} a_{k} x^{k})}^{'} = \sum_{k = 0}^{n} k a_{k} x^{k - 1} \end{matrix}

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Example 1.8.

Let us verify just one instant of the power rule. For

x, x_{0} > 0,

x - x_{0} = (\sqrt{x} + \sqrt{x_{0}}) (\sqrt{x} - \sqrt{x_{0}}),

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\sqrt{x} - \sqrt{x_{0}} = \frac{1}{\sqrt{x} + \sqrt{x_{0}}} (x - x_{0})

The function

φ_{x_{0}} (x) = \frac{1}{\sqrt{x} + \sqrt{x_{0}}}

is continuous at

x_{0} .

Thus, the derivative of

1 / x

x_{0}

- \frac{1}{2 \sqrt{x_{0}}}

and this verifies the power rule in the case

r = 1 / 2 .

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A fundamental operation on functions is composition. One uses the Chain Rule to compute the derivative of a composition.

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Proposition 1.9. The Chain Rule.

Suppose

f (x)

is differentiable at

x = x_{0}

and

g (u)

is differentiable at

f (x_{0})

then the composition

g \circ f (x)

is differentiable at

x_{0} .

Moreover,

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(g \circ f)^{'} (x_{0}) = g^{'} (f (x_{0})) f^{'} (x_{0}) .

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Proof.

It follows from the assumptions that there are functions

γ (u)

and

φ (x)

continuous at

u_{0} = f (x_{0})

and

x_{0},

respective such that

\begin{matrix} (1.4) & f (x) - f (x_{0}) = φ (x) (x - x_{0}) \\ (1.5) & g (u) - g (u_{0}) = γ (u) (u - u_{0}) \end{matrix}

Substituting

u

f (x)

in (1.5) and using (1.4), we get

\begin{aligned} g (f (x)) - g (f (x_{0})) & = γ (f (x)) (f (x) - f (x_{0})) \\ = γ (f (x)) φ (x) (x - x_{0}) \end{aligned}

Since

f (x)

is differentiable at

x_{0},

it is continuous at

x_{0}

as well Proposition 1.5. So, we conclude that the function

x \mapsto γ (f (x)) φ (x)

is continuous at

x_{0}

since continuous function are closed under composition and product. This shows that

g \circ f

is differentiable at

x_{0},

moreover

(g \circ f)^{'} (x_{0}) = γ (f (x_{0})) φ (x_{0}) = g^{'} (f (x_{0})) f^{'} (x_{0}) .

Checkpoint 1.10.

Differentiation alters the parity of a function. That is the derivative of an even (odd) function is odd (even).

Solution.

Let

f (x)

be a differentiable even function. So

f (- x) = f (x)

for all

x

in the domain of

f .

Taking derivative on both sides, yields, according to the Chain Rule,

f^{'} (x) = (f (- x))^{'} = - f^{'} (- x) .

This shows that

f^{'} (x)

is an odd function.

Similarly, if

f (x)

is an odd function, then

- f^{'} (x) = (- f (x))^{'} = (f (- x))^{'} = - f^{'} (- x) .

Thus,

f^{'} (x) = f^{'} (- x)

and so

f^{'} (x)

is even.

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Subsection 1.2.2 Implicit Differentiation

The curve

y^{2} = x

is not the graph of a function of

x

because, for instance, the points

(1, - 1)

and

(1, 1)

on the graph have different

y

coordinates but the same

x

coordinate.

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However, locally around any fixed point

(x_{0}, y_{0})

with

x_{0} \neq 0

the curve is indeed the graph of some function of

x .

This is the simplest case of the Implicit Function Theorem. For example, around

(1, 1)

the curve is the graph of

y = \sqrt{x}

and around

(1, - 1)

it is the graph of

y = - \sqrt{x} .

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Suppose the equation

F (x, y) = 0

defines

y

as a function of

x

locally near a point

(x_{0}, y_{0}) .

Assuming

y (x)

is differentiable at

x = x_{0},

we can try to find the derivative of

y (x)

x = x_{0}

without explicitly writing

y (x)

as an expression of

x

which is often cumbersome, if not impossible to do. Let us illustrate this by our parabola example

y^{2} = x .

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Example 1.12.

Let us compute the slope of the tangent to the curve

y^{2} = x

(or

y^{2} - x = 0

) at the point

(x_{0}, y_{0}) = (1, - 1) .

We do this not by solving

y

explicitly as

y (x) = - \sqrt{x}

then find the derivative at

x = 1

but by differentiating the relation

y^{2} = x .

So we have

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\begin{aligned} \frac{d}{d x} y^{2} & = \frac{d}{d x} x \\ \frac{d}{d y} y^{2} \frac{d y}{d x} & = 1 \\ 2 y \frac{d y}{d x} & = 1 \end{aligned}

Since

(1, - 1)

is on the graph of

y (x),

y (1) = - 1

and so

1 = 2 y (1) y^{'} (1) = 2 (- 1) y^{'} (1) .

That is

y^{'} (1) = - 1 / 2 .

Also, from the relation

2 y (x) y^{'} (x) = 1

we must conclude

y^{'} (x) = 1 / (2 y (x)) .

Therefore, if we know

y

explicitly as

- \sqrt{x},

then we can also express

y^{'} (x)

explicitly as

- 1 / (2 \sqrt{x}) .

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