difference between first and second deriatives?

Question

i don't understand the difference between first and second deriatives.

if the question is 3X^3 + 5X^2 + X -9

and we have to find the stationary points. how would we go about it, using first deriative and second deriative.

details would be appreciated.

Answer 1

y=3x^3 + 5x^2 + x - 9



Get the first derivative



y' = 9x^2 + 10x + 1

Stationary points are where y'=0



9x^2+10x+1=0

(9x+1)(x+1)=0



x1=-1/9, x2=-1



Differentiate y' to get the second derivative

y" = 18x + 10



Let x=x1

y" = -2 + 10 = 8 > 0

This implies that the turning point at (-1/9) is a minimum



Let x=x2

y"=-18+10

= -8 < 0

This implies that the turning point at 1 is a maximum

Answer 2

To find a stationary point, you have to find where the graph is horizontal. The derivative is just the slope, so you have to find where this is zero.



Differentiating 3X^3 + 5X^2 + X -9 gives you 9x^2 + 10x + 1 so the stationary points are the roots of the quadratic equation 9x^2 + 10x + 1 = 0, which are x = –1/9 and x = –1.



So far so good, but you still don't know whether the stationary points are peaks (hills) or troughs (valleys). That's where the second derivative comes in. It gives you the slope of the first derivative.



Why do you need this? Well, at a peak, the second derivative is negative because the first derivative is changing from positive to negative as the original curve slopes up on the left then down on the right. At a trough, the reverse is true.



You find the second derivative simply by differentiating the first derivative. This gives 18x + 10. If you now plug in the values of the stationary points, –1/9 and –1, you get 8 (positive, indicating a trough) and –8 (negative, indicating a peak).



It's well worth the trouble to sketch the graphs of the original function together with its first and second derivatives. Then it will all make much more sense.

Answer 3

you use the first derivative to find the rate of change of the curve, where there is no change in the rate



(ie where dy/dx = 0)



you get the stationary points of the curve.



you then use the second derivative to determine wether the stationary points are maximums or minimums (happy or sad curves) or possible inflections!



i'll go on to solve the equation...



y = 3X^3 + 5X^2 + X -9



dy/dx = 9x^2 + 10x + 1

proceed to factorise this equation

dy/dx = (9x +1)(x + 1)

from this the stationary points are x = -1, x = -1/9



proceed to find out wether the stationary points are max, min or inflex by using the 2nd derivative...



d2y/dx2 = 9x^2 + 10x + 1

d2y/dx2 = 18x + 10



Insert the 2 values of x to give x = 8, x = -8

the stationary point at x = -1/9 is maximum, the stationary point at x = -1 is minumum

Answer 4

The derivative is the slope of the graph.

The second derivative is the rate at which the slope is changing, or the slope of the slope if you wish.



y = 3X^3 + 5X^2 + X -9



dy/dx = 9x^2 + 10x + 1

put this equal to zero to find the values of x where stationary points occur.



d2y/dx2 = 18x + 10

Now put those values of x here to see whether the stationary point is minimum, maximum or inflexion.



Think about hte curve y = x^2.

On the extreme left, the slope is negative and it gradually increases as x increases. As a result the stationary point in this curve is a minimum.

d2y/dx2 = 2 = +ve essentially means that hte slope, dy/dx, is increasing by 2 every time x increases by 1.