Category Archives: Maths

Basic differentiation

Posted on Friday 17th December, 2010 by Joseph

Finding the gradient of a straight line is easy: \frac{\Delta y}{\Delta x}.  Finding the gradient of a curve isn’t as easy.  Drawing a tangent and finding its gradient is one solution, but there’s a better way than that- differentiation.

Differentiating, \frac{dy}{dx}, a function, f(x), results in a derivative function, f'(x):

\displaystyle y = f(x) \rightarrow \frac{dy}{dx} = f'(x) = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x}

To differentiate an entire function, you must differentiate each term individually- and to do this, you follow a simple rule:

\displaystyle y = kx^{n} \rightarrow \frac{dy}{dx} = knx^{n-1}

Example

\displaystyle f(x) = 4x^{5} + 9x^{2} - 6x + x - \sqrt{2x} + \frac{3}{12x^{6}} - 15

You can’t differentiate \frac{3}{12x} and \sqrt{2x} directly, so it’s easier to deal with fractions and other algebraic expressions first and convert them into kx^{n} form:

\displaystyle f(x) = 4x^{5} + 9x^{2} - 6x + x - 2x^{\frac{1}{2}} + 36x^{-6} - 15

Now we can differentiate:

\displaystyle f(x) \rightarrow \frac{dy}{dx} = f'(x) = 20x^{4} + 18x - 6 + 1 - x^{- \frac{1}{2}} - 216x^{-7}

And simplify:

\displaystyle f(x) \rightarrow \frac{dy}{dx} = f'(x) = 20x^{4} + 18x - 5 - x^{- \frac{1}{2}} - 216x^{-7}

And that’s it.

Explanation

But just in case, here’s a brief run-down for each expression at a time…

  • 4x^{5} \rightarrow \frac{dy}{dx} = 20x^{4} because 4 \times 5 = 20.

This is just a straight-forward differentiation here- nothing really to say about it.

  • 9x^{2} \rightarrow \frac{dy}{dx} = 18x because 18x^{1} is just 18x.

Differentiating a quadratic expression leaves x to the power of 1, so there’s no point in writing the index down.

  • -6x \rightarrow \frac{dy}{dx} = -6 because x^{0} = 1;
  • x \rightarrow \frac{dy}{dx} = 1 for the same reason.

When you differentiate x to the power of 1, you create x to the power of 0.  Anything to the power of 0 always equals 1:

\displaystyle y = kx \rightarrow \frac{dy}{dx} = k

  • \sqrt{2x} = 2x^{\frac{1}{2}} \rightarrow \frac{dy}{dx} = x^{- \frac{1}{2}}.

The square root of x can be re-written as x to the power of a half.  Minus 1 from 0.5 and you get -0.5.

  • \frac{3}{12x^{6}} = 36x^{-6} \rightarrow \frac{dy}{dx} = -216^{-7} because 36 \times -6 = -216.

Re-write fractions as negative indices, and then it’s exactly the same as the first expression, only it’s negative this time (both constant and index).

  • -15 \rightarrow \frac{dy}{dx} = 0 because -15x^{1} \rightarrow \frac{dy}{dx} = 0.

When you have only a constant on its own, actually x is to the power of 0- it’s just not shown.  Since you multiply the index by the constant, you’re left with 0:

\displaystyle y = k \rightarrow \frac{dy}{dx} = 0

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Factorising quadratics

Posted on Thursday 21st October, 2010 by Joseph

Simple expressions can be factorised easily, however quadratics have to be factorised into two sets of brackets like this;

ax^{2} \pm bx \pm c = (ax \pm n_{1})(x \pm n_{2})

There’s a simple technique used to factorise a quadratic.  We’ll use the following example:

x^{2} + 7x + 12 = 0

The b and c terms of the quadratic expression should be simple to identify- in our example b=7 and c=12.  The n terms that go in each bracket are factors of c that are also the sum of b.  There are of course several factor pairs of 12:

  • 1 and 12
  • 2 and 6
  • 3 and 4
  • -1 and -12
  • -2 and -6
  • -3 and -4

All of these multiply to give c, but only 3 and 4 add together to give b, 7.

  • 4 \times 3 = 12
  • 4 + 3 = 7

4 and 3 are our factors, so the quadratic factorises to:

x^{2} + 7x + 12 = (x + 3)(x + 4)

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