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Posts Tagged ‘differentiation’

Stationary Points (Maximum and Minimums) and Differentiation

On a graph a stationary point is any point where the gradient is 0 so where the graph is flat. For example the graph y=x2 has one stationary point at the origin.

Finding the Stationary Points

We know that stationary point occur when the gradient is 0 so when the derivative of the graph is 0, so in order to find the stationary points we but first differentiate the curve.

For example lets consider the graph $y = 3x^2 + 2x - 7$. We cab differentiate this to find
$\frac{dy}{dx} = 6x + 2$

We must then equate the derivative to 0 and solve the resulting equation. This is because we are trying to find the points where the gradient is zero and these point occur exactly at the solutions of the equation we have formed.

So in our example we form the equation
$6x + 2 = 0$
by equating our expression for $\frac{dy}{dx}$, $6x + 2$, to 0
Solving this equation we find that stationary points occur exactly when
$x = \frac{2}{6} = \frac{1}{3}$
Note that there can be more than solution to this equation, each of which is a valid stationary point.

Finally we should also find the y co-ordinate for the stationary point by putting this value of x into the initial equation. So for this example $y= 3 \cdot \frac{1}{3}^2 + 2 \cdot \frac{1}{3} - 7 = -6$
So the only stationary point is at $(\frac{1}{3},-6)$

Nature of Stationary Points

The nature of a stationary point simply means what the graph is doing around it and are characterised by the second derivative, $\frac{d^{2}y}{ dx^2}$ (found by differentiating the derivative). There are three types of stationary point:

1. Maximum Points: These are stationary points where the graph is sloping down on either side of the stationary point (a sad face type of curve).
Here ${d^{2}y}{dx^2} < 0$
2. Minimum Points: These are stationary where the graph is sloping upwards on either side of the point (a happy face)

Here ${d^{2}y}{dx^2} > 0$
3. Point of Inflection: Here the direction of the slope of the graph is the same either side of the stationary point, it can be in either direction.

At a point of inflection ${d^{2}y}{dx^2} = 0$ but ${d^{2}y}{dx^2} = 0$ isn’t enough to ensure that a point really is a point of inflection as it could still be a maximum or minimum point
4. Checking the nature of a Stationary Point when ${d^{2}y}{dx^2} = 0$
In this case the easiest thing to do is look a small distance either side of the point and see whether the y value is greater than or less than that of the stationary point. You can then draw yourself a picture to see what it is. For example if they are both greater than the stationary point you know it is a minimum point, but if one is greater and one is less than it is a point of inflection

Warning: checking points either side does not guarantee the correct result as there may be another stationary point or a break in the graph between where you are checking and the stationary point so you should always check using the derivatives if possible

Categories: algebra, calculus

The Chain Rule

The chain rule allows you to differentiate composite functions (functions of other functions) ie) f(g(x)) such as sin(3x2) or (5x3+2x+3)2. The rule is as follows
$\frac{d}{dx}(f(g(x)) = \frac{dg}{dx}\frac{df}{dg}(g(x)) = g'(x)f'(g(x))$
or to understand it more simply you differentiate the inner function and multiply it by the derivative of the outer function (leaving what’s inside alone).

Differentiating brackets raised to a power

The chain rule can be a great short cut to differentiating brackets raised to a power as it doesn’t require you to multiply them all out, it also enables you to differentiate brackets raised to an unknown power.
Consider $\frac{d}{dx}((ax + b)^n)$
This is the composite of the functions ax+b and tn. So we differentiate them both to get a and ntn-1 and then apply the formula to get
$\frac{d}{dx}((ax + b)^n) = an(ax+b)^{n-1}$
Notice how we multiplied the derivative of the inner function, a, by the derivative of the outer function ntn-1 but substituted ax+b back in for t.

To generalise we can replace the ax+b with f(x) and by applying the above get
$\frac{d}{dx}((f(x))^n) = f'(x)n(f(x))^{n-1}$

Differentiating Trigonometric functions

We can also use the chain rule when differentiating sin(f(x)) and cos(f(x)) since we know how to differentiate sin(x) and cos(x).
Using the chain rule we get
$\frac{d}{dx}(sin(f(x)) = f'(x)cos(f(x))$
and
$\frac{d}{dx}(cos(f(x)) = -f'(x)sin(f(x))$

Fundamental Theorem of Calculus

This theorem forms much of the basis of calculus and the uses of differentiation and integration. It basically states that differentiation and integration are opposites so if you differentiate and integral you’ll get the function you started with. This can be stated as follows:

if $F(x) = \int_a(x)^b(x) \! f(t) \, dx$ then $\frac{dF}{dx} = f(a(x))\frac{da}{dx} - f(b(x))\frac{db}{dx}$

or in the more simple case

if $F(x) = \int_0^x \! f(t) \, dx$ then $\frac{dF}{dx} = f(x)- f(0)$

It is this idea that allows us to know, for example,
$\int \! \frac{1}{1+x^2} \, dx = tan^-1(x) + c$
from the knowledge that
\$frac{d(tan^-1(x))}{dx} = \frac{1}{1+x^2}$

This makes much of integration easier as it is often much easier to work out the derivative a function than work out the integral of one so we can look for functions which when differentiated give us the function that we want to integrate and then know that the integral is that function plus a constant.

Categories: calculus, maths

Integrating Fractions – using the natrual logarithm – Example tan(x)

From result found be differentiating the natural logarithm,
$\frac{d}{dx} (ln(f(x))) = \frac{f'(x)}{f(x)}$
for some function f(x),

and the fundamental theorem of calculus we cay say that

$\int \! \frac{f'(x)}{f(x)} \, dx = ln|f(x)| + c$ where c is the integration constant

Simple Example

The most basic example of this is the integration of 1/x,

$\int \! \frac{1}{x} \, dx = ln|x| + c$

More complex example: Integration of tan(x)

A slightly more complicated example of this is the integration of tan(x). To do this we must remember that $tan(x) = \frac{sin(x)}{cos(x)}$ and notice that $\frac{d}{dx}(cos(x)) = -sin(x)$. This means that -tan(x) is of the form $\frac{f'(x)}{f(x)}$ as required. Using this we can get

$\int \! tan(x) \, dx = \int \! \frac{sin(x)}{cos(x)} \, dx = lan|cos(x)| + c$

Trick for using this identity

Sometimes we get integrals that are almost in this form but not exactly, eg) $\int \! \frac{x}{5 + x^2} \, dx$, however to solve these we can often factorise a constant so that it is in the required form. In this example we can take out a 2 so we get $\frac{1}{2} \int \! \frac{2x}{ 5 + x^2} \, dx = ln|5 + x^2| + c$

Categories: calculus

Differentiate Logs with Proof

In order to differentiate logs we must use the chain rule. The simplest type of log to differentiate is a natural log this can be done as shown below.

Differentiate Natural Logs
A natural log is a log to the base e.
d/dx (ln x) = 1/x

However if we want to differentiate ln(f(x)) we must use the chain rule to get

d/dx (ln(f(x)) = f'(x)/f(x)

Proof of Derivative of Natural Logs
Consider

y=ln(x)

then from the definition of a log we get

ey = x                   –(1)

Differentiate each side with respect to x (you need to use implicit differentiation for the left to get ey dy/dx) to get

ey dy/dx = 1

but from (1) we know that ey = x which we can substitute to get

x dy/dx =1

giving the derivative

dy/dx = 1/x

Categories: calculus, maths

Differentiation From First Principles (with example)

When you differentiate a function or curve you are finding the gradient of the tangent to the curve. If the curve you are differentiating a curve or function that is in terms of x, eg y=x2, then the differential is also a curve or function in terms of x, eg 2x. This allows you to find the gradient of the curve for any point on the curve with a known x co-ordinate.

Differentiating from first principles involves finding the gradient of a tangent to a curve from the basic definition of gradient,
grad = (y1 – y2) / (x1-x2)
ie) not by following a rule.

When differentiating from first principles we consider chords (lines passing from one point on the curve to another) from the point on the curve with the x coordinate x to points with a slightly larger x coordinate x+d. The gradient of these chords is an approximation of the gradient of the tangent to the curve at the point with x co-ordinate x and the approximation becomes better as d becomes smaller. If we take the limit as d tends to 0 we find the actual gradient of the tangent

Note: the derivative of a curve y = f(x) is written as dy/dx

Example: y=x2

Consider 2 points P and Q on the curve y=x2 a small distance apart where the difference in their x co-ordinate is d. Then
P(x,x2)
Q(x+d,(x+d)2)
Then the gradient of the chord PQ is given by

To find the derivative we must now take the limit of this as d tends to 0.
So we find

By David Woodford

Categories: calculus, maths

Integration by Parts

July 19, 2009 1 comment

Integration by parts is a method that allows us to integrate the product of two function such as

∫2xe3xdx

where 2x is one function and e3x is another
To do this we use the formula

∫(u d/dx) dx = uv – ∫(v du/dx) dx

where u and v are both functions of x, 2x and e3x in the above example.

Proof of Integration by Parts Formula

This can be shown by considering the product rule for differentiation as shown below
the product rule states that
d(uv)/dx = u dv/dx + v du/dx

If we now integrate both side we get

uv = ∫( u dv/dx) dx + ∫(v du/dx) dx
since integration is the opposite of differentiation

now we can simply rearrange this to get

∫(u d/dx) dx = uv – ∫(v du/dx) dx

Simple Example (without limits)

To demonstrate this formula we shall integrate the example above

∫2xe3xdx

To do this we first need to pick which function (2x or e3x) to allocate to u and which to dv/dx. Picking the correct function is crucial to integration by parts since the method works by allowing us the differentiate the part of the function which is difficult to integrate. In order for this to work it is useful to use a function that will become simpler in some way. For example differentiating e3x gives 3e3x so making the product no easier to integrate, however integrating 2x gives 2 which does make it easier to integrate since we can take the 2 outside as a factor.

So we will let u=2x and dv/dx=e3x
we then differentiate 2x to get 2
and integrate e3x to e3x/3
(it can often be useful to put these in a table with columns u,v and the function on the first row and derivative on the second row)

These can then be put into the formula to find the integral
∫2xe3xdx = 2xe3x/3 – ∫e3x2/3 dx
∫2xe3xdx = 2xe3x/3 – 2e3x/9 + c

Example with Limits

usually, however integration is required to be carried out between limits. To perform integration by parts between limits is quite simple. You takes the value of uv as usual between limits and then set the limits of  ∫v du/dx dx to the limits of the original integral. To demonstrate this we will integrate the above example between 0 and 1

so

10 2x e3x dx = [2xe3x/3]10 – ∫10 e3x2/3 dx
=2e3 – 0 – 2e3/9 + 2/9 = (16e3 + 2) / 9

Categories: calculus, maths