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AS Level - Core 1 – Differentiation Part 1
Differentiating Equations For A Gradient
Differentiation is part of a branch of mathematics called calculus, which is the mathematical
study of change
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Let's say we have a line, given by the equation Y=2X+1
Plotted on a graph it looks like this:

The equation is given in the form Y=mX+c where m is our gradient and c is our Y intercept
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Differentiation is introduced when we start to look at graphs that aren't linear, for example a
graph where the power of X isn't 1
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Y=2X2+2X-2 is a little more complex, because now we have a curve, as shown in the graph
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For starters, the
gradient is constantly changing, so there wouldn't be a single value for it anyway
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This is where the differentiation
comes in
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The notation
for this is given as dY/dX=
This means "The derivative of Y with respect to X" and to calculate it, for each term, we multiply
the coefficient of X by the power of X and then subtract 1 from each power
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We must remember that 2X is the same as 2X1 and that -2 can be written as 2X0
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Now we must subtract 1 from our powers to give us 4X1+2X0-0X-1
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What we might notice here is that any term where the power of X is 1,
we can just throw away the X because the coefficient is unchanged, and any constant, like our –
2, can also be thrown away
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Y=2X2+2X-2 therefore we can calculate that dY/dX=4X+2
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To demonstrate this:
We have our curve Y=2X2+2X-2 and I want to know what the gradient of it will be where X is 2
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Therefore, I can say, the gradient of the curve, when X is 2, is 10
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You could be asked the reverse as well, where you are given a value for a gradient and are
asked to find the co-ordinates of the curve, where a tangent with this gradient would touch
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dY/dX is used to calculate the gradient from an X value, but this time, we
don't know the co-ordinates but do know the gradient, so we're calculating the X value from it
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4X+2=14 so we solve to find X
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Y=2(3)2+2(3)-2
Y=2(9)+6-2
Y=18+4
Y=22
Our co-ordinates for the point on the curve where the gradient is 14 are (3,22)
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Likewise, if it asks you for the co-ordinates, make sure
you find both the X and Y co-ordinates
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Read the question, see
what it is asking for and then answer accordingly
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We then used that to find gradients from given coordinates and vice versa
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As well as searching
for the gradient at a point on a curve, the examiner could ask us to use that gradient to find the
equation of the tangent to the curve at that point
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Y=2X2+2X-2 and we are told we are to find the
equation of the tangent to the curve when the X co-ordinate is 4
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Y-b=m(X-a) where m is the gradient and a and b are co-ordinates
(a,b) of any point the line passes through
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Quite simply, we substitute our X
value into the curve's equation to find Y, like so:
Y=2(4)2+2(4)-2
Y=2(16)+8-2
Y=32+6
Y=38
So the point on the curve our tangent will touch is (4,38)
Now we just need m, the gradient
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We differentiate our
curve's equation and substitute in our X value
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I'm going to give it in the form Y=mX+c
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Sometimes however, instead of being asked for the tangent, we are asked for the normal to the
point on a curve
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It is the line which cuts the curve at a right angle to its
gradient, so is perpendicular to the tangent
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Find the co-ordinates and gradient, then substitute these into the straight line formula
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A gradient perpendicular to another is easy to
find
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We'll do our last example, in which we were asked to find the tangent, but this time, we will be
looking for the normal instead
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We'll first find its negative equivalent, which is –18 and
then we'll invert it
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From then on there is no difference
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This is Differentiation Part 1
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