how exactly does differentiaition work ?? heres my workings
when x = 1 y = 20 (1,20)
when x = 2 y = 30 (2,30)
change in y/change in x = (30-20)/(2-1) = 10/1 = 10
dy/dx = 2x+7
when x = 1 dy/dx = 9
when x = 2 dy/dx = 11
so how do i use these results to understand how differentiation works
eg the gradient at point 1 is 9 and at point 2 = 11
but using my change in y / change in x it says its 10
how does that work many thanks
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how exactly does differentiaition work ?? pos rep for assistance watch
- Thread Starter
- 01-07-2006 10:05
- 01-07-2006 10:26
y=x^2+7x+12 is a quadratic function which has the shape of a curve. Every point on that curve has a different gradient.
When you say that change of y/change in x = 10, you're just calculating the gradient of a line which passes through (1,20) and (2,30), AND is tangent to the curve at a single point.
dy/dx is a function that allows you to calculate the gradient of the tangent of a point.
- 01-07-2006 10:29
Hi, Can i just ask what level of maths are you doing at the moment? Is this your first time doing differentiation?
If it isnt sorry if im patronising you. Differentiation is a method in mathematics and enables us to work out the gradient of a tangent on a curve whether it be a x2 or a x3 etc. This is useful because on these curves the gradients is always changing and when you use differentiation it enables you to find the gradient without manually using tangents to the curve at one specific point.
Later on in differentiation it can be used to find the maximum and minimum points of a curve this is because at a min and max point dy/dx = 0 and you use the 2nd derivate to work out whether its a max or a min.
Also you can differentiation trig functions so you can find the gradient on a sin x, cos x, tan x graph.
That about what i know on differentiation theres more to it than that obviously. Differentiating is also used in say physics with vectors, you can prove an equation of motion by differentiation too.
- 01-07-2006 14:06
At degree level we study the concept of a derivative (the thing we get after differentiation) as a limit, if you are studying A levels then you wont need to know this definition formally, im just giving it to hopefully give you some insight into where differentiation comes from.
The proper formal definition is:
The derivative of a function f(x) is
the limit as h tends to 0 of [f(x+h)-f(x)]/h = dy/dx=f '(x)
the best way to see where this limit comes from is to look at a diagram, the successive straight lines (they are chrords) given by [f(c+h)-f(c)]/h get closer and closer to the tangent line as h gets smaller and smaller (tends to zero) hopefully this link might help a bit more in explaining! (should give diagram too!)
for example to calculate the derivative of f(x)=x^2 from this formal definition of the limit we get dy/dx=limit h tends to 0 [(x+h)^2-x^2]/h
=limit h tends to 0 [x^2+2xh+h^2 - x^2]/h
=limit h tends to 0 [2xh+h^2]/h
=limit h tends to 0 [2x+h]
which is what we get when we apply the normal procedure (not from first principles as I have just done!)
for your question y=x^2+7x+12
dy/dx=2x+7 as you calculated and when x = 1 dy/dx = 9
when x = 2 dy/dx = 11
This is because when you calculated the limit dy/dx the limit comes out as a function which is not constant ie dy/dx=2x+7 so will be different at every point on your original curve, looking at the sketch of the curve we can see the gradient is constantly changing. what you then worked out: change in y/change in x = (30-20)/(2-1) = 10/1 = 10 is not the gradient or rate of change of the curve with respect to x. your second calculation only works in calculating the gradient of a straight line where dy/dx is constant always.
So to find the gradient of a curve at a particular point differentiate as normal and evaluate the derivative at the point your looking for the gradient at.
Really hope this hasnt confused you too much!
- 01-07-2006 15:08
I think my explanation was good enough, but give the rep to Kernel Rev for his effort. Well-deserved.
- 01-07-2006 15:42
You can refer to this website for a 10 mins explanation of the derivative, with voice and animation (click on Continue after each part):
- Thread Starter
- 03-07-2006 14:16
thanks all pos rep for everyone