[james.h]

(Original post by KyraBloke)
Maximum point on a graph is where it turns. As you approach this point, the gradient gets smaller and smaller - i.e reaches 0. Hence why you set it equal to 0. If you set it equal to 1, then at that point, the gradient would be 1, which is larger than 0, hence the gradient there would be bigger than at 0.

To try to clarify what we're doing, let's take a fairly simple example:



Stationary points of y=f(x) are where dy/dx=0. However, if we are looking for the maximum gradient, then we want to take dy/dx as our original function, let's call this g(x):



The maxima and minima of the gradient of f(x) are then found where g'(x)=f''(x)=0, i.e: where sin(x)=0. With reference to the graph of sin(x), this makes sense.

[Mads_300]

(Original post by james.h)
To try to clarify what we're doing, let's take a fairly simple example:



Stationary points of y=f(x) are where dy/dx=0. However, if we are looking for the maximum gradient, then we want to take dy/dx as our original function, let's call this g(x):



The maxima and minima of the gradient of f(x) are then found where g'(x)=f''(x)=0, i.e: where sin(x)=0. With reference to the graph of sin(x), this makes sense.

This is mind blowing I've never really thought about using the second derivative for anything else other than working out nature of stationary points


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[james.h]

(Original post by Mads_300)
This is mind blowing I've never really thought about using the second derivative for anything else other than working out nature of stationary points

Sorry, had to sleep.

In this case, of course, if you wanted to use the "second derivative test" to determine the nature of the stationary points of the gradient, you'd use the second derivative of g(x), which is the third derivative of f(x).

Also, in applications the second derivative can be useful for other things. For instance, the first derivative of displacement is velocity, and the second derivative is acceleration.