Derivatives of trigonometric functions

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Trigonometric rules for differentiation

The derivatives of the fundamental trigonometric functions are shown here.

Function Derivative
\(\sin(x)\) \(\cos(x)\)
\(\cos(x)\) \(-\sin(x)\)
\(\tan(x)\) \(\frac{1}{\cos(x)^2}\)

Let’s practice!

Determine the derivative of the function

\({-5}\cdot\mathrm{sin}\left(x\right)-4\)

We first apply the sum rule:

\(\frac{d}{dx}\left({-5}\cdot\mathrm{sin}\left(x\right)-4\right)=\frac{d}{dx}\left({-5}\cdot\mathrm{sin}\left(x\right)\right)+\frac{d}{dx}\left(-4\right)\)

We now apply the product-with-constant and constant rule:

\(\frac{d}{dx}\left({-5}\cdot\mathrm{sin}\left(x\right)-4\right)={-5}\cdot\frac{d}{dx}\left(\mathrm{sin}\left(x\right)\right)+0\)

The derivative of \(\sin(x)\) is \(\cos(x)\)

\(\frac{d}{dx}\left({-5}\cdot\mathrm{sin}\left(x\right)-4\right)={-5}\cdot\mathrm{cos}\left(x\right)\)

Derivatives of polynomials and power functions

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Here we give the derivative for polynomials and power functions. We start with three handy rules to calculate derivatives.

Three basic rules for differentiation

Let \(c\) be a real number.

  • Constant rule: The derivative of the constant function \(f(x)=c\) is \(f'(x)=0\).
  • Constant multiple rule: The derivative of the product \(c\cdot f(x)\) of the constant \(c\) with a function \(f\) is \(c\cdot f'(x)\).
  • Sum rule: If \(f\) and \(g\) are functions, the derivative of the sum function \(f(x)+g(x)\) is equal to \(f'(x)+g'(x)\).

To differentiate polynomials, we need to get familiar with the derivative of power functions \(x^n\), for which \(n\) is a natural number. This is a special case of the rule below, which covers the derivative of all real power functions.

Power rule for differentiation

If \(a\) is a real number, the derivative of the function \(x^a\) is equal to \(a\cdot x^{a-1}\). In other words: \(\frac{d}{dx}{x}^{a}=a\cdot{x}^{a-1}\)

Using the above-mentioned rule and the sum rule, we can find the derivative of any polynomial.


Let’s practice!

Calculate the derivative of \(y={5}\cdot{x}^{6}\).

Using the polynomial rule for differentiation \(\frac{d}{dx}\left({a}\cdot{x}^{n}\right)={n}\cdot{a}\cdot{x}^{n-1}\) with \(a=5\) and \(n=6\), we find:

\(y^{\prime}=\frac{d}{dx}=\left(5\cdot{x}^{6}\right)=5\cdot\left(6\right)\cdot{x}^{5}=30\cdot{x}^{5}\)

Derivative

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Let \(f\) be a function defined on an interval around a point \(a\). If  \(\begin{array}{rcl}\lim_{h\to 0}\frac{f\left(a+h\right)-f\left(a\right)}{h}\end{array}\)  exists, then \(f\) is called differentiable at \(a\); the limit is called the derivative of \(f\) at \(a\).

This definition is quite intuitive, as it shows how the output of the function changes in proportion to a very small change \(h\) in the input.

This limit is indicated by \({f}^{\prime}{(a)}\) and also by \(\frac{df}{dx}\left(a\right)\). The act of determining the derivative is called differentiation.


If \(f\) is differentiable at all points of an interval \(I\), we say that \(f\) differentiable on \(I\). In that case, \({f}^{\prime}\) is a function on \(I\).

The value \({f}^{\prime}{(x)}\) is often indicated by \(\frac{d}{dx}f\left(x\right)\).

If \(y\) is a function rule of \(f\), we also write \(\frac{d}{dx}y|_{x=a}\) instead of \({f}^{\prime}{(a)}\) or \(\frac{df}{dx}f\left(a\right)\).

The number \({f}^{\prime}{(a)}\) is the slope of the graph of \(f\) at the point \(\left[a,f\left(a\right)\right]\).


Let’s practice!

What is the derivative at \(-4\) of the function \(13x+7\)?

Answer

The graph of the linear function \(13x+7\) is a line, so its slope is equal to the coefficient of \(x\), at any point. Here, that coefficient is equal to \(13\).

You can also use the original definition to find the derivative. Input the function in the definition and take the limit.

\(\begin{array}{rcl}\lim_{h\to 0}\frac{(13x+h)+7-13x+7}{h}\end{array}\)

Multiply out the numerator.

\(\begin{array}{rcl}\lim_{h\to 0}\frac{13x+13x+7-13x+7}{h}\end{array}\)

\(\begin{array}{rcl}\lim_{h\to 0}\frac{13h}{h}=13\end{array}\)

The limit of a constant is the constant itself. We can see that \(h=13\)

Difference quotient

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In order to calculate the slope of a tangent line to a graph, we can take a line that lies close to the tangent line. For this line, you can calculate the slope by using the difference quotient.

Below you see the graph of the function \(f(x)=\frac{1}{5}x^2+6\) and the tangent line \(l\) to \(f\) at the point \(\left[2,\frac{34}{5}\right]\). You can also see the line \(m\) through \(\left[2,\frac{34}{5}\right]\) and \(\left[3,\frac{39}{5}\right]\). Both points lie on the graph of \(f\).
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We can approximate the slope of the tangent line \(l\) by calculating the slope of line \(m\).

The slope of line \(m\) is given by:

\(\begin{array}{rcl}\frac{\text{the difference in y}}{\text{the difference in x}}&=&\frac{\Delta y}{\Delta x}\end{array}\)

\(\begin{array}{rcl}\Delta y=\end{array}\) the difference between the \(y\) coordinates of \(\left[3,\frac{39}{5}\right]\) and \(\left[2,\frac{34}{5}\right]\).

\(\begin{array}{rcl}\Delta x=\end{array}\) the difference between the \(x\) coordinates of the same points.

Therefore:

\(\begin{array}{rcl}\frac{\Delta y}{\Delta x} &=& \frac{\frac{39}{5}-\frac{34}{5}}{3-2} \end{array}\)

We can conclude that

\(\begin{array}{rcl}\frac{\Delta y}{\Delta x} &=& 1 \end{array}\)

We can therefore approximate the slope of the tangent line \(l\) by \(1\).


Such an approximation of the slope of a tangent line of a graph is called a difference quotient. This is so called because it is actually the quotient of two differences \(\Delta y\) and \(\Delta x\) that express the change, \(\Delta x\) in \(x\) and the change, \(\Delta y\) in \(y\), respectively.

For cases where the graph is derived from a function, we can state the following formal definition of the difference quotient:

\(\frac{f\left(a+h\right)-f\left(a\right)}{h}\)

Tangent line

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For various kinds of problems it is useful to know the rate change of a function.

  • In physics, for example, you can calculate the velocity of an object using a graph of position as a function of time.
  • In economics there are many applications of these changes in order to calculate the marginal profit and marginal costs.
  • Often you can use the slope of the graph of a function (which measure the rate of change) in order to calculate the extremes of that function.

One way to capture the change of a function at a point, is the tangent line.


Tangent line

Consider a graph containing a point \(P\). A line \(l\) through \(P\) is a tangent line at \(P\) to the graph if:

  • in a small neighborhood around \(P\), the point \(P\) is the only point that the line \(l\) has in common with the graph;
  • in a small neighborhood around \(P\), the points of the line \(l\) all lie on the same side of the graph.

The tangent line at \(P\) to a graph need not exist and need not be unique. If it is unique, then the slope of the tangent line \(l\) is a good measure of the rate of change of the graph at the point \(P\) and we can consider the slope of the graph at that point \(P\).


Below you see the graph \(f(x)=x^2-x\). Additionally, the lines \(a\), \(b\), \(c\), and \(d\) are drawn.

Which of these lines is/are tangent lines of the graph \(f(x)\)?

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Line \(a\) and line \(d\) are tangent lines to the graph.

Line \(a\) is a tangent line at \(x=-1.58\): the line has only that point in common with the graph, and lies entirely below the graph around that point.

Line \(d\) is a tangent line at \(x=\frac{1}{2}\): the line has only that point in common with the graph, and lies entirely below the graph around that point.

Line \(b\) and line \(c\) are not tangent lines, because they do not have a small interval in which there is only one point of intersection with the graph and in which all other points are on one side of the graph.

Quadratic equations

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Quadratic polynomials and equations

An expression in the form \(a\cdot x^2+b\cdot x+c\), for which \(a\), \(b\) and \(c\) are real numbers is called a quadratic polynomial if \(a\ne0\). It is called quadratic after the highest power of \(x\) occuring in the expression.

The equation \(a\cdot x^2+b\cdot x+c=0\) is called a quadratic equation.


Solution of a quadratic equation by factorization

Consider the quadratic equation with unknown variable \(x\):

\(a\cdot x^2+b\cdot x+c=0\tiny,\)

in which \(a\), \(b\) and \(c\) are real numbers with \(a\ne0\).

If the left-hand side can be factorized in linear factors:\( a\cdot x^2+b\cdot x+c=(x-p)\cdot(x-q)\tiny,\) in which \(p\) and \(q\) are real numbers, the solution of the equation is \(x=p\lor x=q\).

We use the following rule here:

\(A\cdot B=0\phantom{xxx}\) is equivalent to \(\phantom{xxx}A=0\lor B=0\), in which \(A = x-p\) and \(B = x-q\).


Example

Let’s factorize to solve \(x\) in the equation

\(x^2-3\cdot x+1=x+1\)

We can start by reducing the equation to \(0\):

\(x^2-4\cdot x=0\)

We can now factorize the left-hand side:

\((x-4)\cdot x=0\)

This means that

\(x-4=0\vee x=0\),

since \(A\cdot B=0\) if, and only if \(A=0\lor B=0\).
We can now move the constant terms to the right-hand side to get to the solution:

 \(x=0\vee x=4\)

Solving linear inequalities by reduction

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Solve by reduction of linear inequalities with a single unknown

Linear inequalities with a single unknown can always be solved by reduction: step-by-step simplification of the inequality, in which the same operation is performed on both the left-hand side and the right-hand side.

  • Subtracting the same term on both sides is such an operation. For example, \(2x+3\ge5\) can be reduced to \(2x\ge2\) by subtracting \(3\) on both sides.
  • Another such operation is dividing both sides by the same constant \(a\) not equal to \(0\). If \(a>0\), the inequality sign remains. If \(a<0\), the inequality sign has to be reversed in order to get an equivalent inequality. For example, \(2x \ge 9\) can be reduced to the solution \(x\ge\frac{9}{2}\) by dividing both sides by \(2\). \(-2x \ge 9\) can be reduced to the solution \(x\le-\frac{9}{2}\) by dividing both sides by \(-2\).
  • Similar terms can always be brought together. For example, \(2x+3+x+9 < 7x\) can be simplified to \(3x+12 <7x\). We call such a step a simplification.

We perform these operations to get from an equation like \(2x+3+x+9 < 7x\) to an equation with the form \(x>3\), because if we do so we have determined that the solution consists of all \(x\) with a value greater than \(3\).


We will solve the inequality as \(2x+3+x+9 < 7x\) an example.

\(\begin{array}{rcl}a\cdot x+b\ge0&\Leftrightarrow&a\cdot x\ge -b\end{array}\)

We’ll start by simplifying the left hand side:

\(\begin{array}{rcl}3x+12 &<& 7x\end{array}\)

We can now subtract \(7x\) on both sides:

\(\begin{array}{rcl}-4x+12 &=& 0\end{array}\)

Let’s subtract \(12\) on both sides:

\(\begin{array}{rcl}-4x&<& -12\end{array}\)

Finally, we divide both sides by \(-4\):

\(\begin{array}{rcl}x &>& 3\end{array}\)

The solution is therefore \(x>3\).

 

Introduction to solving linear inequalities

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Linear inequalities with one unknown can be solved by reduction. Before we discuss this, we will introduce two logical symbols which are handy when handling systems of linear inequalities (and equations).

Composition of inequalities

The logical symbols \(\land\) and \(\lor\), which mean ‘and’ and ‘or’ respectively, can be used to compose multiple inequalities.

In this way, we can use \(x\ge 2 \land x < 5\) to denote that \(x\) is larger or equal to \(2\), and smaller than \(5\). This can also be shortened to \(2 \le x < 5\).


We can use the notion of equivalence to denote that two inequalities have the same solution, similar to when we’re working with equalities.

Equivalence for inequalities

A system of inequalities is equivalent to another system if they have the exact same solutions. For example: \(x> 5\land x> 2\) is equivalent to \(x> 5\).

We can denote that two expressions are equivalent by using \(\Leftrightarrow\) in between them. For example:

\(x> 5\land x> 2\Leftrightarrow x> 5\).


The following inequalities are all equivalent to one another:

  • \(x> 5\)
  • \(x-5>0\)
  • \(5< x\)
  • \(-x< -5\)
  • \(-5> -x\)

The inequality \(x^2>0\) has \(x\ne0\) as a solution, or \(x<0\lor x>0\).

  • The inequality \(0>0\) with unknown \(x\) does not have a solution.
  • For the inequality \(1>0\) with unknown \(x\), any value of \(x\) is a solution.

What are linear equations?

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An equation with unknown \(x\) is called lineair if the only terms it consists of are constants and constant multiples of \(x\).

Let’s start with some examples:

\(\begin{array}{rcl}x&=&7\\ 2x &=&9\\ 2x+3&=&5\\ 2x+3&=&7x+5\\ 2x+3+x+9 &=& 7x\\ 3x+2&=&0\\ 3&=&0\\ 0&=&0\end{array}\)


In more theoretical words, a linear equation looks like this:

\(a+b\cdot x + c\cdot x+d+\cdots = p+q\cdot x + r + s\cdot x+\cdots\tiny,\)

in which \(a\), \(b\), \(c\), \(d\), \(p\), \(q\), \(r\), and \(s\) are real numbers.


The first example,  \(\begin{array}{rcl}x&=&7\end{array}\), not only is an equation, but it is also the solution to the equation: the only value of \(x\) for which the equation is true, is \(7\).

In general, equations can be rewritten in such a way that the left-hand side is a linear function of #x# and the right-hand side is equal to #0#.

The following two examples are exceptions:

\(\begin{array}{rcl}3&=&0\\ 0&=&0\end{array}\)

They are equations without terms \(x\). For ease, we will call these linear equations too, even though we do not have a linear function of \(x\).

  • The equation \(3=0\) with unknown \(x\) does not have a solution: Equality is not true for any value of \(x\).
  • Every value of \(x\) is a solution to the equation \(0=0\) with unknown \(x\). Equality holds true anyway.

Powers and exponents

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Power of integers

For each number \(g\) not equal to \(0\) and each positive integer (whole number) \(n\) the following applies

\(\begin{array}{rcl} g^n & =& \underbrace{g \times g \times \cdots \times g}_{n\textrm{ times}}\end{array}\)

Here \(g^n\) , the \(n\) power of \(g\), is determined for each integer \(n\). The number \(g\) is called the base and \(n\) is called the exponent.

Here are two  more rules regarding exponents.

\(\begin{array}{rcl}g^0 & =& 1 \\ g^{-n} & = &\frac{1}{g^n}\end{array}\)

Calculation rules for powers

Here are some quick and easy rules to calculate with exponentations!

\(\begin{array}{rcl} \left(\frac{a}{b}\right)^n&=&\frac{a^n}{b^n}\\ \left(\frac{a}{b}\right)^{-n}&=&\frac{b^n}{a^n}\\ \left({a}\cdot{b}\right)^{n}&=&{a^n}\cdot{b^n}\\ {a}^{m}\cdot a^{n}&=&a^{m+n}\\ \left({a}^{m}\right)^{n}&=&a^{m\cdot n}\end{array}\)

These rules apply when \(a\) and \(b\) are numbers not equal to \(0\) and \(m\) and \(n\) are integers.