Archive

Archive for the ‘Differentiation’ Category

Sketching derivative graphs (cubics)

December 1, 2010 Leave a comment

The important thing to remember here is that we are doing a sketch.  It does not have to be accurate, but id does have to conatian all the key points.

Hints:

  1. A cubic differentiates to a quadratic (parabola)
  2. A the turning points have a gradient of zero and will move to the x axis on the derivative sketch
  3. A point with a positive gradient will be above the x axis in the sketch
  4. A point with a pnegative gradient will be below the x axis in the sketch

Examples

Describe and sketch the derivative graph for the function below

The critical points are x=0\mbox{ and }x=\frac{4}{3}, because this is where the turning points occur.

These points will be on the x axis in the sketch.

For x<0 the gradient is positive.

These points will be above the x axis in the sketch.

For 0<x<\frac{4}{3} the gradient is negative .

These points will be below the x axis in the sketch.

The graph is a cubic so the sketch will be a parabola

Practice

Categories: Differentiation

Questions: Differentiation

November 7, 2010 Leave a comment
  1. Find the coordinates of the point where the curve y=x^2+2x-4 has a gradient of 4.
  2. Find the gradient of the curve y=\frac{x^2}{2}-3x+7 at x=3.
  3. Find the x coordinate where the curves 3x^2+2x-7 and 2x^2-5x have the same gradient.
  4. Find the gradients at the points of intersection of the two curves x^2+3x-5 and x^2+2x-1
  5. Find the coordinates of the points where the curve \frac{x^3}{3}+2x^2-21x+4 has a gradient of 0.
  6. Find the gradient of the curve with equation 2x^2-5x+3 at the point (3,6)
  7. Differentiate x^3+3x and hence show that the curve y=x^3+3x never has a gradient of zero.
  8. Show that the curve y=(x-1)^2(3-x) can be written in the form y=3-7x+5x^2-x^3 and find the gradient at (0,3)
  9. Find the gradient of the curve y=5x^2+6x-5 at the point x=1
  10. Find the three gradients where the curve y=(2x+1)(4x-1) cuts the axes
Categories: Differentiation

Finding the turning points of a polynomial

October 29, 2010 Leave a comment

The turning points are sometimes called the stationary points or maximums and minimums.

Lets look at what happens to the gradient at a turning point

  • If we consider the maximum, you can see that before it has a positive gradient and after it has a negative gradient.  During this transition it has a gradient of zero.  So at a maximum \frac{dy}{dx}=0
  • If we consider the minimum, you can see that before it has a negative gradient and after it has a positive gradient.  During this transition it has a gradient of zero.  So at a minimum \frac{dy}{dx}=0
  • So we can see that at a turning point (maximum or minimum) \frac{dy}{dx}=0

Examples

Find the maximum of the curve y=15-2x-x^2

  1. We need to find the gradient function \frac{dy}{dx}
  2. We let \frac{dy}{dx}=0, because this is when the maximum will occur
  3. We solve the equation to find the value of x where the maximum occurs
  4. We substitute this value in the original equation to get the value for y

\frac{dy}{dx}=-2-2x

Let \frac{dy}{dx}=0 \Rightarrow -2-2x=0 \Rightarrow x=-1

If x=-1 \Rightarrow y = 15-2(-1)-(-1)^2 = 16

\therefore the maximum value is y=16

Find the coordinates of the turning points of y=x^3-6x^2+9x

  1. We need to find the gradient function \frac{dy}{dx}
  2. We let \frac{dy}{dx}=0, because this is when the turning points will occur
  3. We solve the equation to find the values of x where the turning points occurs
  4. We substitute the values into the original equation to get the values for y

\frac{dy}{dx}=3x^2-12x+9

Let \frac{dy}{dx}=0 \Rightarrow 3x^2-12x+9=0 \Rightarrow 3(x^2-4x+3)=0 \Rightarrow 3(x-1)(x-3)=0

So x=1 of x=3

If x=1 then y=4 and if x=3 then y=-18

So the coordinates of the turning points are (1,4) and (3,-18)

Given that the curve y=x^2+ax-5 has a stationary point at x = 4 find the value of a

  1. We need to find the gradient function \frac{dy}{dx}
  2. We let \frac{dy}{dx}=0, because this is when the turning points will occur and we will let x = 4
  3. We will solve the resulting equation to find the value of a

\frac{dy}{dx}=2x+a

when x=4, \frac{dy}{dx}=0 \Rightarrow 0=8+a \Rightarrow a=-8

 Find the equation of the tangent to the curve y=2x^3+3x-5 at x=1

  1. We need to find the gradient function \frac{dy}{dx}, because to get the equation of a line we need the gradient and a point it passes through
  2. Substitute x=1 into the equation for y to get the required point the line passes through
  3. Substitute x=1 into the equation for \frac{dy}{dx} to get the gradient of the line
  4. Use y-y_1=m(x-x_1)  to get the equation of the line

\frac{dy}{dx}=6x^2+3

At x=1, y=0

At x=1, \frac{dy}{dx}=9

So substituting into y-y_1=m(x-x_1) gives y=9(x-1)

Categories: Differentiation

Differentiating Polynomials from first principles

October 28, 2010 Leave a comment

The gradient of a curve is always changing.  To calculate the gradient at a point we can consider the gradient of a chord going through that point and gradually make the length of the chord shorter.  As the length gets closer to zero the gradient of the chord should get closer to the gradient of the tangent at the point.

When we are finding the gradient function we are differentiating

Consider the diagram


You can see from the graph that as \delta x gets smaller so does \delta y and that the gradient of the chord gets closer to that of the tangent.

So as \delta x \rightarrow 0, \frac{\delta y}{\delta x} \rightarrow the gradient at x

We refer to the gradient at x as \frac{dy}{dx}

Finding the gradient function of x^2

Let y=x^2

The point that is \delta x across from x has the coordinates  (x+\delta x, y+\delta y)

Substituting into y=x^2 gives y+\delta y=(x+\delta x)^2 \Rightarrow y+\delta y=x^2+2x(\delta x)+(\delta x)^2

Since  y=x^2 we get x^2+\delta y=x^2+2x(\delta x)+(\delta x)^2 \Rightarrow \delta y=2x(\delta x)+(\delta x)^2

If we now divide  through by \delta x we get \frac{\delta y}{\delta x} = 2x+\delta x

Now if we let \delta x \rightarrow 0 we get \frac{dy}{dx}=2x

Finding the gradient function of x^n

Note that (x+a)^n = x^n + nax^{n-1} + \frac{n(n-1)}{2!}a^2x^{n-2} + \frac{n(n-1)(n-2)}{3!}a^3x^{n-3} + ...

Let y = x^n

The point that is \delta x across from x has the coordinates  (x+\delta x, y+\delta y)

Substituting into y=x^n gives y+\delta y=(x+\delta x)^n

So y+\delta y = x^n + n\delta x x^{n-1} + \frac{n(n-1)}{2!}(\delta x)^2x^{n-2} + \frac{n(n-1)(n-2)}{3!}(\delta x)^3x^{n-3} + ...

Since  y=x^n

we get x^n+\delta y = x^n + n\delta x x^{n-1} + \frac{n(n-1)}{2!}(\delta x)^2x^{n-2} + \frac{n(n-1)(n-2)}{3!}(\delta x)^3x^{n-3} + ...

If we now divide  through by \delta x

we get \frac{\delta y}{\delta x} = n x^{n-1} + \frac{n(n-1)}{2!}\delta xx^{n-2} + \frac{n(n-1)(n-2)}{3!}(\delta x)^2x^{n-3} + ...

Now if we let \delta x \rightarrow 0 we get \frac{dy}{dx}=n x^{n-1}

You should be able to see that when we differentiate that we multiply by the power and then subtract one from the power.

Examples

y \frac{dy}{dx}
x^5 5x^4
3x^7 21x^6
5x^{-3} -15x^{-4}

Questions

Categories: Differentiation