Derivative of a function

We begin our study of differential calculus by revisiting the notion of secant lines to a given function f passing through the points (x₀, f(x₀)) and (x₀ + h, f(x₀ + h)). The slope of this line is given by an equation in the form of a difference quotient:

$$m_{\text{sec}}=\frac{f(x_0+h)-f(x_0)}{h}$$

with h ≠ 0.

Fig.1 When Q tends to P, then the chord PQ becomes a tangent at P, as h ⟶ 0.

The Figure shows that as the values of h get closer to 0, the secant lines also approach the tangent line. The slope of the tangent at point x₀ of f is the limit as h approaches 0 of the slope of the secant. The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications across many disciplines. These applications include velocity and acceleration in physics. This limit occurs so frequently that we give this value a special name: the derivative. The process of finding a derivative is called differentiation. We give the following definition.

Definition 5.1.1. (Derivative of a function) Let f: [a,b] ⟶ ℝ. The derivative of f at x₀ ∈ (a,b) is, denoted by f'(x₀) and defined by

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x_0+h)-f(x_0)}{h}$$

provided this limit exists.  □

The line of equation:

is called tangent line to the graph of f at the point (x₀, f(x₀)).

We know from theorem Theorem 4.1.8 that a limit exists, if and only if both the limits, the left hand and the right hand, exist and are equal. Then as in the case of one-sided continuity, the right derivative of f at x and the left-derivative at x, are defined to be the limits

$$f_{+}^{\prime }(x)=\underset{h\to 0^{+}}{lim}\frac{f(x+h)-f(x)}{h}\text{and}{f}_{-}^{\prime }(x)=\underset{h\to 0^{-}}{lim}\frac{f(x+h)-f(x)}{h}$$

provided that these limit exist, and are equal, the function f is differentiable at x.

Definition 5.1.2. We say that f: [a, b] → ℝ, is differentiable on [a,b] if f'(x) exists on (a,b) and bot f'₊(a) and f'(b) exist.  □

When f is differentiable in every point (a,b), is defined the function f': (a, b) → ℝ, known as derivative of f given by

Differentiation of elementary functions

The derivative of a constant

The derivative of a constant function is zero, indeed if f(x) ≡ k we have ∀x:

thus for h → 0

The graph of f equals the horizontal line y = k, the slope of the tangent at each point of the graph is zero, and hence all tangents coincide with the graph of f itself.

Identity function

Let f(x) = x in some open interval I. For every x ∈ I,

Hence the derivative of the identity function is everywhere equal to 1, and the tangent at a point (x,x) of the graph of f equals the straight line y = x. Thus, the tangent line again is simply the original line itself.

We now prove that the derivative of f(x) = x² is equal to 2x:

$$\frac{f(x+h)-f(x)}{h}=\frac{(x+h{)}^2-x^2}{h}=\frac{x^2+2hx+h^2-x^2}{h}=2x+h$$

taking the limit for h → 0, we have 2x.  ■

Trinometric functions

Evaluate the derivative of the sine function:

$$\frac{\sin (x+h)-\sin x}{h}=\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}=\sin x\frac{\cos h-1}{h}+\frac{\sin h}{h}\cos x\underset{h\to 0}{\overset{}{\to }}\cos x$$

for h → 0 recalling remarkable limits, we have cos x. ■

The following brief table lists derivatives of some of the elementary functions

f	f'
k (constant)	0
x	1
x2	2x
1/x	−1/x2
√x	1/2√x (for x > 0)
xα (α ∈ ℝ, x > 0)	αxα−1
sin x	cos x
cos x	−sin x
ex	ex
log x	1/x
tg x	1 + tg2x = 1/(cos x)2
cotg x	−(1 + cotg2x) = −1/sin2 x
Sh x	Ch x
arcsin x	1/sqrt(1 − x2)
arcos x	−1/sqrt(1 − x2)
arctg x	1/sqrt(1 + x2)

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