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Derivative of a function

Geometrically, the problem of finding the derivative of the function is existence of the unique tangent line at some point of the graph of the function. The problem of finding the unique tangent line at some point of the graph of the function is equivalent to finding the slope of the tangent line at the same point.

Let s be a secant line trough points P(p, f(p)) and Q(q, f(q)). A secant line trough these two points has the slope:

    \[m_{sec} = \frac{f(q) - f(p)}{q-p}.\]

If q \to p, that is, if point Q according to the graph of the function approaches point P, then the  secant line s becomes the tangent line t. Therefore, the slope of the secant line m_{sec} becomes the slope of the tangent line m_{tan}, that is

    \[m_{tan}= \lim_{q \to p} \frac{f(q) - f(p)}{q-p}.\]

The expression \frac{f(x) - f(x_o)}{x - x_0} is called the difference quotient for the function f at point x_0.

 

Let I \subseteq \mathbb{R} be an open interval. The function f: I \to \mathbb{R} is differentiable at the point x_0 \in I if exists

    \[\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}.\]

The number

    \[\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}\]

is called the derivative of the function f at point x_0 and we write

    \[f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}.\]

The function f is differentiable on I if it is differentiable at every point x_o \in I.

Therefore, the equation of a tangent at point (x_0, f(x_0)) is

    \[y - f(x_0) = f'(x_0) (x - x_0).\]

To find a tangent  to the curve at point (x_0, f(x_0)), it’s enough to find its slope, and that number is f'(x_0) (provided that this limit exists).

If we let \Delta x = x - x_0, then x= \Delta x + x_0 and \Delta x \to 0 as x \to x_0, the definition of the derivative of the function f:  I \to \mathbb{R} at point x_0 we can also write in the following form:

    \[f'(x_0) = \lim_{\Delta x \to 0 } \frac{f( x_0 + \Delta x) - f(x_0)}{\Delta x},\]

where \Delta x is the change in x-coordinates and \Delta y = \Delta f(x) = f(x_0 + \Delta x) - f(x_0) is the change in y-coordinates.

 

A derivative can also be written with symbols:

    \[f'(x) = \frac{d f(x)}{dx} = \frac{d y}{d x}.\]

 

Theorem.

Let I \subseteq \mathbb{R} be an open interval and f: I \to \mathbb{R}. If the function f is differentiable on I, then the function is continuous on I.

The opposite is not true. A counterexample; the absolute value function abs: \mathbb{R} \to \mathbb{R}^+ defined as

    \[abs(x) = \begin{cases} x, & \mbox{if } x \ge 0 \\ -x, & \mbox{if } x<0 \\ \end{cases}\]

is continuous on \mathbb{R}, however, it doesn’t have a derivative at point x = 0.

Derivative rules

Let I \subseteq \mathbb be an open interval and f, g: I \to \mathbb{R} differentiable functions at point x_o \in I. Then

1.) the function f + g is differentiable at point x_0 and

    \[(f+g)' = f' + g'\]

,

2.) (the Leibniz rule) the function f \cdot g is differentiable at point x_o and

    \[f \cdot g = f' \cdot g + f \cdot g'\]

,

3.) the function \alpha \cdot f, \alpha \in \mathbb{R} is differentiable at point x_0 and

    \[(\alpha \cdot f)' = \alpha \cdot f'\]

,

4.) the function \frac{f}{g} defined on I is differentiable at point x_0 and

    \[\left( \frac{f}{g}\right)' = \frac{f' \cdot g - f \cdot g' }{g^2}\]

.

 

The derivative of a constant function f: \mathbb{R} \to \mathbb{R}, f(x) = c at x_0 \in \mathbb{R} is equal to

    \[f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = \lim_{x \to x_0}\frac{c - c}{x - x_o} =0,\]

that is, f'(x) = 0.

The derivative of a function f: \mathbb{R} \to \mathbb{R}, f(x) = x at point x_0 \in \mathbb{R} is equal to

    \[f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = \lim_{x \to x_0}\frac{x - x_0}{x - x_0} =\lim_{x \to x_0} 1 = 1 ,\]

that is, f'(x) = 1.

For all n \in \mathbb{Z}, n\neq 0 is valid (x^n)' = n \cdot x ^{n-1}. For n < 0 the potential x^n we can write as a fraction:

    \[x^n = \frac{1}{x^{-n}} = \frac{1}{x^t},\]

where t = -n is a positive number.

That is, we have:

    \[(x^n)' = \left ( \frac{1}{n} \right)' = - \frac{tx^{t-1}}{x^{2t}} = -t \frac{1}{x^{t+1}} = n \frac{1}{x^{-n+1}} = nx^{n-1}.\]

Example 1. Find the derivative of function f if

    \[f(x) = 3 x^3 -2x^2 + 3x -1.\]

Solution:

    \[\begin{aligned} f'(x) &= (3 x ^3) ' - (2x^2)' +(3x)' - 1' \\ & = 3 \cdot 3 \cdot x ^{3 - 1} - 2 \cdot 2 \cdot x ^{2 - 1} + 3 \cdot 1 - 0 \\ & = 9 x^2 - 4 x + 3. \\ \end{aligned}\]

 

Example 2. Find the derivative of function f if

    \[f(x) =(x^2 - 1) ( 3x +4)\]

Solution:

    \[\begin{aligned} f'(x) &= (x^2 - 1)' \cdot (3x +4) + (x^2 - 1) (3x + 4)' \\ &= 2x \cdot (3x + 4) + (x^2 - 1) \cdot 3 \\ &= 6x^2 + 8x + 3x^2 -3 \\ &= 9x^2 + 8x -3 .\\ \end{aligned}\]

Example 3. Find the derivative of function f if

    \[f(x) = \frac{x^2 - 6x + 5}{x-3}\]

Solution:

    \[\begin{aligned} f'(x) &= \frac{(x^2 - 6x + 5)' \cdot (x-3) - (x^2 - 6x + 5) \cdot (x-3)'}{(x-3)^2} \\ & = \frac{(2x - 6) \cdot (x-3) - (x^2 - 6x + 5) \cdot 1}{(x-3)^2} \\ &= \frac{2x^2 - 6x - 6x + 18 - x^2 + 6x -5}{(x+3)^2} \\ &= \frac{x^2 - 6x +15}{(x+3)^2} \\ \end{aligned}\]

Derivatives of trigonometric functions

Derivative of function sine.

We will use the trigonometric identity for difference of sines. Firstly, we calculate the difference quotient:

    \[\begin{aligned} \frac{f(x) - f(x_0)}{x - x_0} &= \frac{\sin x - \sin x_0}{x - x_0}\\ & = \frac{2 \sin \frac{x - x_0}{2} \cos \frac{x + x_0}{2}}{x - x_0} \\ & = \frac{\sin \frac{x - x_0}{2}}{\frac{x - x_0}{2}} \cos \frac{x + x_0}{2}. \\ \end{aligned}\]

By using the definition of derivative we have:

    \[\begin{aligned} f'(x_0) &= \lim_{x \to x_0} \frac{\sin \frac{x - x_0}{2}}{\frac{x - x_0}{2}} \cos \frac{x + x_0}{2}\\ & = \lim_{x \to x_0} \frac{\sin \frac{x - x_0}{2}}{\frac{x - x_0}{2}} \cdot \lim_{x \to x_0} \cos \frac{x + x_0}{2}. \\ \end{aligned}\]

Since \lim_{x \to 0} \frac{\sin x}{x} = 1, then

    \[\begin{aligned} f'(x_0) &= 1 \cdot \lim_{x \to x_0}\cos \frac{x+ x_0}{2} \\ & = \cos \frac{2 x_0}{2} \\ & = \cos x_0.\\ \end{aligned}\]

That is, we obtained (\sin x)' = \cos x.

Similarly, we can find derivatives of the rest trigonometric functions:

 

 

 

 

Higher order derivative

If the function f: I \to \mathbb{R} has the derivative at every point of open interval I, then the function f' also has the derivative on I. The derivative of the function f' is denoted as f'' and is called the second derivative of the function f.

The nth derivative of the function f is denoted as f^{(n)} and defined as

    \[f^{(n)}(x) = \left (f^{(n-1)} (x) \right)'.\]

 

Example 4. Determine the third derivative of the function f if:

    \[f(x) =2 x^5 - 3x.\]

Solution:

    \[f'(x) = 2 \cdot 5 \cdot x^{5-1} - 3 = 10x^4 -3\]

    \[f''(x) = (f'(x))' = (10 x^4 - 3) ' = 40 x^3\]

    \[f'''(x) = (f''(x))' = (40 x^3)' = 120 x^2\]

The composite function rule

If f and g are differentiable functions, then a composite function f(g(x)) is differentiable and the derivative of a composite function f(g(x)) we calculate by the following formula:

    \[f(g(x))' = f'(g(x)) \cdot g'(x).\]

The formula above is also known as the chain rule.

Example 5. Calculate the derivative of the following function:

    \[f(x) = \sqrt{1 - 2x}.\]

Solution:

Firstly, note that f(x) = f(g(x)), where f(x) = \sqrt{x} and g(x) = 1 - 2x.

Therefore, we have:

    \[f'(x) = \frac{1}{2 \sqrt{x}} \Longrightarrow f'(g(x)) =\frac{1}{2 \sqrt{1 - 2x}}\]

and

    \[g'(x) = (1 -2x)' = -2,\]

Now, by the chain rule, we have:

    \[f'(x) = \frac{1}{2 \sqrt{1 - 2x}} \cdot (- 2) = -\frac{1}{\sqrt{1 - 2x}}.\]

 

Derivative of inverse function

Let f be injective function and f'(x) \neq 0. Then its inverse function f^{-1} is differentiable at point y=f(x) and

    \[(f^{-1})'(y) = \frac{1}{f'(x)}, \quad \quad f'(x) = \frac{1}{(f^{-1})'(y)}.\]

For instance,  y = \sqrt{x} \Longleftrightarrow x = y^2, that is, the function f^{-1}(y) = y^2 is an inverse function of function f(x) = \sqrt{x}. By using the formula f'(x) = \frac{1}{(f^{-1})'(y)} we calculate:

    \[(\sqrt{x})' = f'(x) = \frac{1}{(f^{-1})'(y)} = \frac{1}{2y} = \frac{1}{2 \sqrt{x}}.\]

 

Derivative of logarithmic function

All logarithmic functions can be represented as logarithmic function of one base. Therefore, we choose f(x) = \ln x, which base is the number e.

Note.  The function \ln: \mathbb{R}^+ \to \mathbb{R} defined as f(x) = \ln x is called the natural logarithm function.

The number e is defined as the limit of a sequence:

    \[e = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n.\]

 

Firstly, we calculate the difference quotient:

    \[\frac{f(x + \Delta x) - f(x)}{\Delta x} &= \frac{\ln (x  + \Delta x) - \ln x}{\Delta x}.\]

Now, by using the rule of logarithms \left ( \log x - \log y = \log \frac{x}{y} \right) we have:

    \[\begin{aligned} \frac{f(x + \Delta x) - f(x)}{\Delta x} &= \frac{1}{\Delta x} (\ln (x + \Delta x) - \ln x) \\ &= \frac{1}{\Delta x} \ln \left (\frac{x + \Delta x}{x} \right ). \end{aligned}\]

By the definition of derivative now we have:

    \[\begin{aligned} (\ln x)' &= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \ln \left (\frac{x + \Delta x}{x} \right ) \\ & =\lim_{\Delta x \to 0} \frac{1}{\Delta x} \ln \left ( 1 +\frac{\Delta x}{x} \right ). \end{aligned}\]

If we define n: = \left| \frac{x}{\Delta x} \left | then n \to +\infty when \Delta x \to 0. Since the domain of the function \ln is \mathbb{R}^+, then n> 0, that is x >0. Therefore, we can write n = \frac{x}{\Delta x}. It follows that \frac{\Delta x}{x} = \frac{1}{n} and \Delta x = \frac{x}{n}, that is \frac{1}{\Delta x} = \frac{n}{x}.

By the substitution we obtain:

    \[(\ln x)' = \lim_{n \to + \infty} \frac{n}{x} \ln \left( 1 + \frac{1}{n} \right) .\]

By using the logarithm rule \log_a x^n = n \log_a x we have:

    \[(\ln x)' = \lim_{n \to + \infty} \frac{1}{x} \ln \left(1 + \frac{1}{n \right)^n} = \frac{1}{x} \lim_{n \to + \infty}\ln \left(1 + \frac{1}{n \right)^n} .\]

We defined e = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n, therefore

    \[(\ln x)' = \frac{1}{x} \cdot \ln e,\]

that is,

    \[(\ln x)' = \frac{1}{x}.\]

Derivative of logarithm of any other base we calculate from the connection of logarithm functions:

    \[\log_a x = \frac{1}{ \ln a} \cdot \ln x\]

Therefore,

    \[(\log_a x)' = \frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}.\]

Derivative of exponential function

The exponential function is an inverse function of logarithmic function:

    \[y = e^x \Longleftrightarrow x = \ln y.\]

By using the rule for derivative of inverse function we have:

    \[(e^x)' = \frac{1}{(\ln y)'} = \frac{1}{\frac{1}{y}} = y = e^x.\]

A derivative of the exponential function a^x we calculate from the connection:

    \[a^x = e ^{x \ln a} \Longrightarrow (a^x)' = e^{x \ln a}  \cdot (x \ln a)' = a^x \cdot \ln a.\]

 

 

Note;

    \[(\ln f) ' = (\ln f)' \cdot f' = \frac{1}{f} \cdot f',\]

    \[(e^f)' = (e^f)' \cdot f' = e^f \cdot f'.\]

 

Example 6. Find the derivative of function f if

    \[f(x) = \ln ( \sin x).\]

Solution:

 

    \[f'(x) =\frac{1}{\sin x} \cdot (\sin x)' = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x.\]

Example 7. Calculate the first derivative of function f at point x = 1 if

    \[f(x) = e^{x-2}.\]

Solution:

    \[f'(x) = (e^{x-2})' = e^{x-2} \cdot (x -2)' = e^{x-2} \cdot (-2).\]

It follows

    \[f'(1) = (-2) \cdot e^{1-2} = (-2) \cdot e^{-1} = -\frac{2}{e}.\]

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