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Calculus Introduction to Differentiation graph : slope : derivative : differentiation : dy/dx

Differentiation and The Derivative

Introduction

• Differentiation (finding derivatives of functions)

• Integration (finding indefinite integrals or evaluating definite integrals)

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.

The derivative is given the symbol

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

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

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

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

Constants and Powers of x

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.

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

Using the formula for the derivative (dy/dx = anx^n-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)

When y = -2x³, dy/dx = -6x².

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

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

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.

(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

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.

(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

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.

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

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

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

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.

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

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.

Implicit Differentiation

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

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.

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

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

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

This is simply given by dy/dx = 1/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:

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

Ln(2^y) = Ln(3^Sin(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

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.

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

y = e^x is the only function equal to its own derivative.

Uses of the Derivative

Solving Equations (Newton's Method of Approximations)

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

Newton's formula for approximations is given by:

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

• Take the function and put the value x = a. This is f(a).
• Take the derivative of the function, F(x) and put the value x = a. This is F(a).
• Divide the two values: f(a) / F(a).
• Subtract this from the guess, a.
• This will give an approximate value for x for the original equation, f(x) = 0.

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.

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:

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

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:

• a = 3
• f(a) = f(3) = 3² - 10 = -1
• F(a) = F(3) = 2 × 3 = 6

Using Newton's formula:

x_approx = 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.

• a = 3.1666
• f(a) = f(3.1666) = 3.1666² - 10 = 0.0273
• F(a) = F(3.1666) = 2 × 3.1666 = 6.3332

Using Newton's formula for the second time gives:

x_approx = 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.

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.

• a = 2
• f(a) = f(2) = 2³ - (5 × 2) + 3 = 1
• F(a) = F(2) = (3 × 2²) - 5 = 7

Using Newton's formula:

x_approx = 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:

• a = 1.857
• f(a) = f(1.857) = 1.857³ - (5 × 1.857) + 3 = 0.1187
• F(a) = F(1.857) = (3 × 1.857²) - 5 = 5.3453

Using Newton's formula (second time):

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

Using this value:

• a = 1.834
• f(a) = f(1.834) = 1.834³ - (5 × 1.834) + 3 = -0.0012
• F(a) = F(1.834) = (3 × 1.834²) - 5 = 5.0906

Using Newton's formula (third time):

x_approx = 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.

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.

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.

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.

• a = 1
• f(a) = f(1) = Cos(1) - 1 = -0.4596
• F(a) = F(1) = -Sin(1) - 1 = -1.8414

Using Newton's formula:

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

Using this value:

• a = 0.7504
• f(a) = f(0.7504) = Cos(0.7504) - 0.7504 = -0.0189
• F(a) = F(0.7504) = -Sin(0.7504) - 1 = -1.6819

Using Newton's formula for the second time:

x_approx = 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:

• a = 0.7391
• f(a) = f(0.7391) = Cos(0.7391) - 0.7391 = -0.0000 (to four decimals)
• F(a) = F(0.7391) = -Sin(0.7391) - 1 = -1.6735

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

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.

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.

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.

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.

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