Assuming function of one variable...
Want y[x] s.t. y''[x] =y[x]
The characteristic equation is r^2 = 1.
r = (+/-) 1
So, e^(x*1) and e^(x * -1) work. To get full generality, multiply them by any constants A,B and add.
y[x] = A e^x + B e ^(-x)
y'[x] = A e^x - B e ^(-x)
y''[x] = A e^x + B e ^(-x) = y[x]
When the first derivative of the function is equal to zero and the second derivative is positive.
Take the derivative of the function and set it equal to zero. The solution(s) are your critical points.
A derivative graph tracks the slope of a function.
That sounds a lot like a critical point to me.
Any monomial in the format: axn has a derivative equal to: nax(n - 1) In this case, "a" is equal to 1 and "n" is equal to 2. So the derivative of x2 is equal to 2x.
When the first derivative of the function is equal to zero and the second derivative is positive.
All it means to take the second derivative is to take the derivative of a function twice. For example, say you start with the function y=x2+2x The first derivative would be 2x+2 But when you take the derivative the first derivative you get the second derivative which would be 2
you have to first find the derivative of the original function. You then make the derivative equal to zero and solve for x.
well, the second derivative is the derivative of the first derivative. so, the 2nd derivative of a function's indefinite integral is the derivative of the derivative of the function's indefinite integral. the derivative of a function's indefinite integral is the function, so the 2nd derivative of a function's indefinite integral is the derivative of the function.
The derivative refers to the rate at which a function changes with respect to another measure. The differential refers to the actual change in a function across a parameter. The differential of a function is equal to its derivative multiplied by the differential of the independent variable . The derivative of a function is the the LIMIT of the ratio of the increment of a function to the increment of the independent variable as the latter tends to zero.
Set the first derivative of the function equal to zero, and solve for the variable.
The signum function is differentiable with derivative 0 everywhere except at 0, where it is not differentiable in the ordinary sense. However, but under the generalised notion of differentiation in distribution theory, the derivative of the signum function is two times the Dirac delta function or twice the unit impulse function.
The function is called the signum function, or sign(x). It is equal to abs(x)/x
The "critical points" of a function are the points at which the derivative equals zero or the derivative is undefined. To find the critical points, you first find the derivative of the function. You then set that derivative equal to zero. Any values at which the derivative equals zero are "critical points". You then determine if the derivative is ever undefined at a point (for example, because the denominator of a fraction is equal to zero at that point). Any such points are also called "critical points". In essence, the critical points are the relative minima or maxima of a function (where the graph of the function reverses direction) and can be easily determined by visually examining the graph.
You didn't specify the equation. A minimum or maximum value of a function is often found by calculating the derivative of a function, writing an equation for derivative equal to zero, and then analyzing points where the derivative either doesn't exist, or is equal to zero. You'll find find information about this in introductory calculus books.
That means that either the function is equal to zero everywhere (y = 0), or it is the exponential function (y = ex).
The same way you get the second derivative from any function. Assuming you have a function that expresses potential energy as a function of time, or perhaps as a function of position, you take the derivate of this function. This will give you another function. Then, you take the derivate of this derivative, to get the second derivative.