y=3x4-8x3-3
Let's say we want to calculate dy/dx. For a term Axn, the differential with respect to x is Anxn-1. For a constant value, we can say that n=0 because x0=1. This means that when you differentiate a constant (e.g. -3) with respect to x, it "disappears".
Therefore,dy/dx = (34)x3-(83)x2 = 12x3-24x2
It may be helpful to think of dy/dx as a gradient of a line or a curve on your typical x-y axis graph. For a curve with a complicated expression (e.g. y=x2-5), the value of the gradient is different at different points, depending on the value of x. This is where the concept of a "tangent" comes in handy.
You might be asked to find the second differential with respect to x, d2y/dx2. For this, we just differentiate dy/dx again.d2y/dx2 = (123)x2-(242)x = 36x2-48x
If dy/dx is the gradient, d2y/dx2 is how much the gradient changes for a change in x.
You might be asked to find a stationary point. A stationary point is a point on your curve where the gradient (dy/dx) is equal to 0. We want to find the x values that stationary points occur at, so we set dy/dx equal to 0.
12x3-24x2=0Since both the terms include an x raised to the power 2 or higher, we can take out x2 as a factor.
x2(12x-24)=0
This means we have either x2=0, or 12x-24=0. Both are valid solutions are both should be written down.
x2=0therefore, a stationary point occurs at x=0.
12x-24=0x-2=0x=2therefore, a stationary point also occurs at x=2.
You might also be asked to determine the nature of a stationary point, whether it is a minimum, maximum, or point of inflexion. The rule for calculating this using the second derivative, is:d2y/dx2>0, it is a minimum pointd2y/dx2<0, it is a maximum pointd2y/dx2=0, it can be a minimum, maximum, or point of inflexion
Let's find out what is the nature of the stationary point at x=2.d2y/dx2 = 36x2-48xSubstitute in x=2d2y/dx2=36(2)2-48(2)= 36(4)-96=144 - 96 = 48
At x=2, d2y/dx2>0, and therefore the stationary point is a minimum.