Differentiation is a mathematical tool that evaluates the way a function changes with respect to some independent variable. Essentially, the derivative of a function at a specific point is the instantaneous slope of the function at that point. A function that is at a maximum has a positive slope prior to the maximum, and a negative slope after the maximum. That means, in nearly all cases, the derivative of the function is zero at the maximum. We can use that fact to identify local minima and maxima of any continuous, differentiable function.
Finding Minimum and Maximum Derivatives
Find the derivative of your function.
If your function, f(x)=3x, then your derivative, f'(x)=3.
If g(y)=4(y-2)^2 + 6, then your derivative, g'(y)=8*(y-2).
If h(z)=sin(z), then h'(z)=cos(z).
Find the derivative of the derivative of your function, otherwise known as the second derivative.
From the examples:
For f(x)=3x, and f'(x)=3, then f''(x)=0.
For g(y)=4(y-2)^2 + 6, and g'(y)=8*(y-2), then g''(y)=8.
For h(z)=sin(z), and h'(z)=cos(z), then h''(z)=-sin(z).
Set the second derivative equal to zero. The second derivative of your function will be equal to zero only when the first derivative has a minimum or maximum.
Each of the three examples above demonstrates different behavior. For f(x)=3x, f''(x)=0. For what values of x is f''=0? All of them. Therefore, your derivative has a minimum or maximum at every point, which doesn't make sense until you remember that the derivative, f'(x) is equal to 3 everywhere. So it has no minima or maxima, or it has the same maximum and minimum everywhere, which is 3.
For g(y)=4(y-2)^2 + 6, g''(y)=8. For what values of y is g''=0? None of them; it's always equal to 8, so the derivative of your function has no minima or maxima. Again, it seems strange until you look at the graph and see that your initial quadratic function g(y) has a first derivative that's just a straight line---no dips or bumps to make extrema.
For h(z)=sin(z), h''(z)=-sin(z). For what values of z is -sin(z)=0? At z=0, +/-pi, +/-2*pi, etc. Now look back at the first derivative and plug in the values of z that we now believe to correspond to minima and maxima. h'(z)=cos(z). Cos(0)=1, which we know is a maximum for the cosine function. Cos(pi)=-1, which we know is a minimum for cosine, etc.
Now restrict the range for your independent variable to find the relative maximum and minimum derivatives. In this context, relative maximum just means the maximum over a given range of independent variables. In our third example above, we could ask for the relative maximum between z= 3pi and 5pi, and we'd find extrema at 3pi, 4pi, and 5*pi. For this example, the cosine function is well known to the point where we know that it's minimum at 3 pi and 5 pi, and maximum at 4 pi.
This step has given us the extrema, but it doesn't tell us for sure which are maxima and which are minima. One final step will clear up the remaining confusion.
Take the derivative of your function one more time. If it's positive at the extremum, then it's at a minimum, if it's negative, you're at a maximum.
Our example again: the second derivative is h''(z)=-sin(z), the derivative of that is h'''(z)=-cos(z). In the range z=3pi to 5pi, the second derivative was equal to zero at 3pi, 4pi, and 5pi, so those are the values we're interested in. -cos(3pi)=1, which is positive, so the extrema we found is a minimum. -cos(4pi)=-1, so the extrema is a maximum. And cos(5pi)=1, so the extrema there is another minimum. All that is consistent with what we know of the cosine function.
Tips & Warnings
- As with all math problems, the beauty and the pitfalls are in the details: write your steps out and exercise caution.
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