Calculus

Introduction to Differentiation

Slope of Curve

graph : slope : derivative : differentiation : dy/dx

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Differentiation and The Derivative

Introduction

Calculus is a very important branch of mathematics. It is a form of mathematics applied to continuous graphs (graphs without gaps). Calculus has two aspects:

The calculus was invented by European mathematicians, Isaac Newton and Gottfried Leibnitz.

This essay introduces Differentiation.

The derivative allows us to calculate the slope or tangent of a graph at any point, P. The process by which a derivative is found is called Differentiation.

The graph below is a simple parabola whose equation is y = x2.

Slope

The derivative is given the symbol

dy/dx

This is pronounced d y by d x or dy dx.

The derivative is a function that gives the slope (tangent) of the graph at any point. The derivative measures the rate of change of y with respect to x. It describes in precise mathematical terms how y changes when x changes. This concept is very important in science.

It can be shown that if y = x2, then the derivative is given by

dy/dx = 2x

So for this curve, when x = 1, the slope is 2; the slope at x = 3 is 6.

The derivative of y = x3 is found to be dy/dx = 3x2.
For y = x4, the derivative is dy/dx = 4x3.

Example 1: Find the slope of the curve y = x3 at the points x = -1 and x = 2 given that the derivative is 3x2.

The derivative of this curve is dy/dx = 3x2.

When x = -1, dy/dx = 3; when x = 2, dy/dx = 12.


Derivatives of Common Functions

In this section I will list the rules for finding derivatives of common types of functions.

Constants and Powers of x

Function Derivative
y = a dy/dx = 0
y = axn dy/dx = anxn-1
This formula works for all values of n (a and n are numbers).

The derivative of a number is zero.

axn is a function consisting of a number (a) multiplied by x raised to a power, n. To find the derivative of this function, multiply the number by the power (an) and reduce the index power by 1.

Example 2: Find the derivatives of the following functions:

(i) y = 3x3; (ii) y = 1/x; (iii) y = 2√x.

Using the formula for the derivative (dy/dx = anxn-1) we can show that

(i) when y = 3x3, dy/dx = 9x2
(ii) when y = 1/x, this can be written as y = x-1. Therefore dy/dx = -x-2 = -1/(x2)
(iii) when y = 2√x, this can be written as y = 2x(1/2). Therefore dy/dx = x-(1/2) = 1/(√x)

Example 3: Find the slope of the graph y = -2x3 when x = -2.

When y = -2x3, dy/dx = -6x2.

For x = -2, dy/dx = -24.

Example 4: Show that the curves y = 3x and y = 2 have constant slopes.

y = 3x can be written y = 3x1; dy/dx = 3x0 = 3 which is a constant. The "curve", y = 3x is a straight line with a slope of 3.

y = 2 can be written y = 2x0, dy/dx = 0 (another constant). The "curve" y = 2 is a straight line parallel to the x axis (a zero slope).

Addition and Subtraction

This is how to differentiate functions that are added together or subtracted.

Function Derivative
y = u + v dy/dx = du/dx + dv/dx
y = u - v dy/dx = du/dx - dv/dx
In these equations, u and v are both functions of x.

If two functions are added together, they can be differentiated separately and the derivatives added. If two functions are subtracted, they can be differentiated separately and the derivatives subtracted.

Example 5: Differentiate the equations (i) y = 4x2 + 2x + 3 and (ii) y = x5 - 5/x.

(i) For the function y = 4x2 + 2x + 3, the derivative is dy/dx = 8x + 2

(ii) For the function x5 - 5/x, the drivative is dy/dx = 5x4 + 5/x2

Sine and Cosine

This is how to find the derivatives of sines and cosines.

Function Derivative
y = aSin(u) dy/dx = a(du/dx)Cos(u)
y = aCos(u) dy/dx = -a(du/dx)Sin(u)
In these equations, u is a function of x. (a is a number).
The value of u (or x) must be in radians.

The derivative of a sine is the cosine. Multiply the resulting cosine by the derivative of the function inside the original sine. The derivative of a cosine is minus the sine. Multiply the resulting sine by the derivative of the function inside the original cosine.

Example 6: Find the derivatives of: (i) y = Sin(x), (ii) y = 3Cos(2x), (iii) y = Sin(x2), (iv) y = x - Cos(x).

(i) y = Sin(x); dy/dx = Cos(x)

(ii) y = 3Cos(2x); dy/dx = -6Sin(2x)

(iii) y = Sin(x2); dy/dx = 2xCos(x2)

(iv) y = x - Cos(x); y = 1 + Sin(x)

Products

Products are functions multiplied together.

Function Derivative
y = uv dy/dx = u (dv/dx) + v (du/dx)
In this equation, u and v are functions of x multiplied together.
This is called the differentiation of products.

If two functions are multiplied, the derivative is found as follows. The first function is multiplied by the derivative of the second function. The second function is multiplied by the derivative of the first function. These two new terms are added together.

Example 7: Differentiate y = xSin(x)

This is a product (uv) so we use the above formula for differentiating products.

dy/dx = xCos(x) + Sin(x).

Example 8: Find the derivative of y = (x2 + 1)√x3

y = (x2 + 1)√x3 can be written y = (x2 + 1)x(3/2)

Using the formula for differentiating products,

dy/dx = (x2 + 1)(3/2)x(1/2) + x(3/2)(2x) = (3/2)(x2 + 1)√x + 2x√x3

Quotients

These are a pair of functions divided.

Function Derivative
y = u/v (v(du/dx) - u(dv/dx)) / v2
In this equation, u and v are functions of x in the form of a division.
This is called the differentiation of quotients.

If two functions are divided, the derivative is found as follows. The denominator function (the one below the line) is multiplied by the derivative of the numerator function (the one above the line). The numerator function is multiplied by the derivative of the denominator function. These two new terms are subtracted together and divided by the square of the original denominator.

Example 9: Differentiate y = Tan(x).

y = Tan(x) can be written as y = Sin(x) / Cos(x). This is a quotient.

dy/dx = [Cos(x).Cos(x) - Sin(x).-Sin(x)] / Cos2(x), (using the above quotient formula)

= [Cos(x).Cos(x) + Sin(x).Sin(x)] / Cos2(x) = [Cos2(x) + Sin2(x)] / Cos2(x)

= 1 / Cos2(x)

The reciprocal of a cosine is called a secant (Sec): 1 / Cos(x) = Sec(x). Therefore the derivative of y = Tan(x) is

dy/dx = Sec2(x).

(Remember that Cos2(x) + Sin2(x) = 1)

Example 10: Find the slope when x = 0 of the curve, y = Sin(x) / (x2 + 1).

Using the quotient formula on y = Sin(x) / (x2 + 1),

dy/dx = [(x2 + 1)Cos(x) - 2xSin(x)] / [(x2 + 1)2]

When x = 0, dy/dx = (0 + 1 - 0) / (0 + 1)2 = 1 / 1 = 1.

(Remember that Cos(0) = 1)

Implicit Differentiation

This allows us to differentiate functions that contain y's mixed with the x's.

Function Derivative
(Function in y) (Derivative of y)(dy/dx)
The y in a function can be differentiated, as long as it is multiplied by dy/dx.
This is called implicit differentiation.

If there is a function in y it can still be differentiated. Differentiate it as before then multiply by dy/dx.

Example 11: Find the slope of the circle with equation x2 + y2 = 4 at the point (0, -2).

This equation could be solved for y and then differentiated. But is simpler to use implicit differentiation:

2x + 2y(dy/dx) = 0.

Rearranging gives: dy/dx = -2x/2y = - x/y

At the point x = 0, y = -2, dy/dx = 0.

Example 12: Express dy/dx in terms of x for the equation Sin(y) = x2.

Differentiating implicitly gives (dy/dx)Cos(y) = 2x. Rearranging to (dy/dx) = 2x/Cos(y).

Remembering that Cos2(y) + Sin2(y) = 1, we can say that Cos(y) = √[1 - Sin2(y)].

Substituting gives (dy/dx) = 2x/√[1 - Sin2(y)] = 2x/√[1 - x4].

Inverse Trigonometric Functions

An expression like Sin(y) = x can be rewritten as y = ArcSin(x), where the expression ArcSin(x) is the called the inverse sine of x. The inverse trigonometric functions have their own rules for differentiation.

Function Derivative
y = ArcCos(x) dy/dx = -1 / [1 - x2]1/2
y = ArcSin(x) dy/dx = 1 / [1 - x2]1/2
y = ArcTan(x) dy/dx = 1 / [1 + x2]
The angles (x) must be in Radians.

Example 13: Find the value of dy/dx for the equation y = ArcCos(x) when x = 0.5.

The derivative is given by dy/dx = -1/[1 - x2]1/2.

Putting the value of x = 0.5 gives dy/dx = -1/[1 - 0.52]1/2 = -1.154.

Logarithms

Function Derivative
Ln(u) (1/u)(du/dx)
In this equation, u is a function of x.
Ln is the natural logarithm (to base e).

The derivative of the natural logarithm of a function is the reciprocal of the function multiplied by the derivative of the function.

Example 14: Find dy/dx for the equation y = Ln(x).

This is simply given by dy/dx = 1/x.

Example 15: Differentiate y = Ln(Cos(x)).

Using the above formula gives dy/dx = (1/Cos(x)).-Sin(x) = -Sin(x)/Cos(x) = -Tan(x)

Logarithms can be used to differentiate more complex functions:

Example 16: Find dy/dx when 2y = 3Sin(x).

2y = 3Sin(x) cannot be differentiated as it is. We can take logarithms on both sides:

Ln(2y) = Ln(3Sin(x)).

Remembering the logarithmic rules of indices, we can rewrite this as:

yLn(2) = Sin(x)Ln(3). This can now be differentiated implicitly:

(dy/dx)Ln(2) = Ln(3)Cos(x) which gives dy/dx = Ln(3)Cos(x)/Ln(2).

Exponential Functions

These functions contain the variable as an index.

Function Derivative
au (du/dx)auLn(a)
eu (du/dx)eu
In this equation, u is a function of x, a is a number.
The number e is the base of natural logarithms.

The derivative of a number raised to the power of a function is the number raised to the function multiplied by the derivative of the function multiplied by the log of the number. If the number is e the derivative of the function is simply multiplied by e raised to the function.

Example 17: Differentiate the following function: y = 23x.

Using the above formula for y = 23x gives dy/dx = 3.Ln(2).23x.

Example 18: Differentiate the following functions: (i) y = eSin(x), (ii) y = ex.

(i) Using the above formula for y = eSin(x) gives dy/dx = Cos(x).eSin(x).

(ii) The derivative of y = ex is dy/dx = ex.

y = ex is the only function equal to its own derivative.


Uses of the Derivative

Solving Equations (Newton's Method of Approximations)

Calculus can be used to obtain approximate solutions to equations. This can be used for calculating roots and for getting values for equations that cannot be easilly solved algebraically.

Before giving the formula and the method, I will define the following shorthand terms:

SymbolMeaningExamples
f(x)
A function of x x2 - 3x + 5 or
Sin (3x) - 2x
f(a)
The value of the function of x, f(x), when x has been set to a If f(x) = x2 - 3x + 5,
f(0) = 5 (set x to 0 in f(x)),
f(1) = 3 (set x to 1 in f(x))
F(x)
The derivative of the function of x, f(x) If f(x) = x2 - 3x + 5,
F(x) = 2x - 3
(the derivative)
F(a)
The value of the derivative of the function of x, F(x), when x = a If f(x) = x2 - 3x + 5,
F(x) = 2x - 3 and F(1) = -1

To solve (find the value of x) an equation in the form,

f(x) = 0

Newton's formula for approximations is given by:

xapprox = a - f(a) / F(a)

where f(x) is the function to be solved and a is a guess at the solution.

This is the method used with the formula:

This value for x is based on the value chosen for a. The better the original guess for a, the closer that x will be to the correct value. If the guess, a, is not close to the correct value, this formula may not work at all.

The new value of x can then be inserted into the formula and the process repeated until the desired accuracy is reached. The closer the guessed value (a) is to the correct value, the less times the formula needs to be used.

A repeating process like this is called iteration.

Some examples will show how the formula works.

Example 19: Find a value for √10 to three decimal places.

This means solving the equation x2 = 10 which can be rearranged to x2 - 10 = 0. This is now in the desired format, f(x) = 0. The function, f(x), has the value:

f(x) = x2 - 10

The derivative of the function, F(x) is therefore:

F(x) = 2x

By looking at the equation ("by inspection"), we know that the solution to this equation is close to 3 (because 32 = 9) so we set the value of the first guess, a, to 3. We can then write down the componnets needed for using Newton's formula:

Using Newton's formula:

xapprox = a - f(a) / F(a) = 3 - f(3) / F(3) = 3 - (-1 / 6) = 3 + 1/6 = 3.1666

We began by guessing that x was 3 and have ended up with a better approximation (3.1666). We can now use this new value in the formula to get an even better approximation.

Using Newton's formula for the second time gives:

xapprox = a - f(a) / F(a) = 3.1666 - f(3.1666) / F(3.1666) = 3.1666 - (0.0273 / 6.3332) = 3.1622

Repeating the process with the new value gives a third value of x as 3.1623 to four decimal places. After using the formula just three times the answer comes out as x = 3.162 to three decimal places.

Historical Note: Newton's Approximations is a general version of a rule used by the ancient Babylonians to find square roots of numbers.

Example 20: Find a positive value for x that satisfies the cubic equation, x3 - 5x + 3 = 0

We begin by writing the function and its derivative: f(x) = x3 - 5x + 3 and F(x) = 3x2 - 5

By looking at the equation, we can see that f(1) = -1 and f(2) = 1 so there must be a value close to 2 which will give f(x) = 0. We can put our guess value (a) equal to 2; the actual value of x will be slightly less.

Using Newton's formula:

xapprox = a - f(a) / F(a) = 2 - f(2) / F(2) = 2 - (1 / 7) = 1.857

This means that 1.857 is a closer approximation to the value of x than 2 was. We can now set a to 1.857 and run the process again:

Using Newton's formula (second time):

xapprox = a - f(a) / F(a) = 1.857 - f(1.857) / F(1.857) = 1.857 - (0.1187 / 5.3453) = 1.834

Using this value:

Using Newton's formula (third time):

xapprox = a - f(a) / F(a) = 1.834 - f(1.834) / F(1.834) = 1.834 - (-0.0012 / 5.0906) = 1.834

So by the third iteration, the value has settled (to three decimal places) to x = 1.834.

Note that for this cubic function, there is another positive value of x between 0 and 1. By putting a to 0, and running Newton's formula three times, it is possible to obtain a second value of x which is close to 0.656.

Cubic equations normally have three roots. There is another value of x that can be found by setting the guess (a) to a value of -2. The reader will be pleased to know that I will leave that as an exercise.

Example 20: Solve the equation Cos(x) = x to three decimal places.

There is no simple method of solving this equation algebraically. We could do it graphically by plotting the graphs of y = Cos(x) and y = x on the same piece of paper and finding the x value of where they intersect. This is shown in the diagram below.

y = x and y = Cos(x)

By rearranging, we have Cos (x) - x = 0 so we can write the function and its derivative as

f(x) = Cos(x) - x and F(x) = -Sin(x) - 1, respectively

From our knowledge of the Cosine, we know that its value oscillates between y = 1 and y = -1 for all values of x. We also know from our knowledge of straight line graphs that y = x is a straight line with a positive slope running through the origin. From this analysis (and by inspecting the graph above), we can infer that the two functions will meet in one place close to the value, x = 1. We can therefore set our first guess (a) to 1.

As a reminder, when trigonometric functions are differentiated we must work in Radians rather than degrees. So we can now evaluate the components for Newton's formula.

Using Newton's formula:

xapprox = a - f(a) / F(a) = 1 - f(1) / F(1) = 1 - (-0.4596 / -1.8414) = 0.7504

Using this value:

Using Newton's formula for the second time:

xapprox = a - f(a) / F(a) = 0.7504 - f(0.7504) / F(0.7504) = 0.7504 - (-0.0189 / -1.6819) = 0.7391

Using this new value:

From the above it can be seen that the value of f(0.7391) is zero to four decimals so there will not be any change to the value of the approximation.

Therefore, to three decimals, x = 0.739, after three iterations.

Rates of Change

The derivative measures rates of change for functions that are continuous and variable. Functions like this are used extensively in science.

If there is a relationship between distance traveled (s) and time (t), then the derivative of distance with respect to time, ds/dt, gives the velocity (v) at any time.

Example 21: A particle moves so that its distance (in m) from a fixed point is given by
s = 2t2 - 3t + 1, where t is time in seconds. Find its velocity after 4s.

The velocity, v, is given by ds/dt so we differentiate the above equation with respect to t:

v = ds/dt = 4t - 3. When t = 4s, v = 13m/s.

If there is a relationship between the velocity of a particle (v) and time (t), then the derivative of v with respect to t, dv/dt, gives the acceleration (a) at any time.

Example 22: The above particle's velocity is given by v = 4t - 3. What it is acceleration after 1s.

The acceleration, a, is given by dv/dt so we differentiate the above equation with respect to t:

a = dv/dt = 4. The acceleration is constant at 4m/s2.

If there is a relationship between energy (E) and time (t), then the derivative of E with respect to t, dE/dt, gives the power (P) at any time.

Example 23: A device uses up energy in a manner dependent on time: E = t3, where E is energy in Joules and t is the time in seconds. Find the power being used after 2s.

The power, P, is given by dE/dt so we differentiate the above equation with respect to t:

P = dE/dt = 3t2. The power after 2s is therefore 12W.

Differentiation is thus one of the most powerful tools in mathematics and physics.

© 2001, 2009 KryssTal


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