The derivative of a curve is basically the slope of the curve. If we say, for example, that if y = 2x, the derivative is 2, that means that at any point the line has this slope. If we say that for the function y = x2, the derivative is 2x, that means that at any point "x", the slope is twice the value of "x".
The formula for the derivative of an inverse (finv)' = 1/(f' o (finv)) allows you get a formula for the derivative of the inverse of any function that you already know the derivative of. For example: What is the derivative of sqrt(x)? You could figure this out using the definition of the derivative, but it is complicated. You already know that the derivative of x2 is 2x. So let f = x2; finv = sqrt(x), f' = 2x. This gives: (sqrt(x))' = 1/(2 sqrt(x)). Now you have derived a "square root rule" with almost no work.
Express the cosecant in terms of sines and cosines; in this case, csc x = 1 / sin x. This can also be written as (sin x)-1. Remember that the derivative of sin x is cos x, and use either the formula for the derivative of a quotient (using the first expression), or the formula for the derivative of a power (using the second expression).
Write sec x as a function of sines and cosines (in this case, sec x = 1 / cos x). Then use the division formula to take the first derivative. Take the derivative of the first derivative to get the second derivative. Reminder: the derivative of sin x is cos x; the derivative of cos x is - sin x.
Use the formula for the derivative of a power. The square root of (x-5) is the same as (x-5)1/2.
If by "2aXaXa", you actually mean "2a3", then the derivative with respect to a is 6a2. On the other hand, if you actually mean "2a3X2", then it's derivative with respect to X would be 6a2X2(da/dx) + 4a3X. If "a" is simply a constant though, then it's derivative is 4a3X
The formula for the derivative of an inverse (finv)' = 1/(f' o (finv)) allows you get a formula for the derivative of the inverse of any function that you already know the derivative of. For example: What is the derivative of sqrt(x)? You could figure this out using the definition of the derivative, but it is complicated. You already know that the derivative of x2 is 2x. So let f = x2; finv = sqrt(x), f' = 2x. This gives: (sqrt(x))' = 1/(2 sqrt(x)). Now you have derived a "square root rule" with almost no work.
Express the cosecant in terms of sines and cosines; in this case, csc x = 1 / sin x. This can also be written as (sin x)-1. Remember that the derivative of sin x is cos x, and use either the formula for the derivative of a quotient (using the first expression), or the formula for the derivative of a power (using the second expression).
Write sec x as a function of sines and cosines (in this case, sec x = 1 / cos x). Then use the division formula to take the first derivative. Take the derivative of the first derivative to get the second derivative. Reminder: the derivative of sin x is cos x; the derivative of cos x is - sin x.
Finding the derivative. The derivative is the measure of how a function changes as its input changes.
Use the formula for the derivative of a power. The square root of (x-5) is the same as (x-5)1/2.
If by "2aXaXa", you actually mean "2a3", then the derivative with respect to a is 6a2. On the other hand, if you actually mean "2a3X2", then it's derivative with respect to X would be 6a2X2(da/dx) + 4a3X. If "a" is simply a constant though, then it's derivative is 4a3X
The derivative at any point in a curve is equal to the slope of the line tangent to the curve at that point. Doing it in terms of the actual expression of the curve, find the derivative of the curve, then plug the x-value of the point into the derivative to find the derivative at that point.
Find the derivative of Y and then divide that by the derivative of A
I am assuming the you are talking about the graph of the derivative. The graph of the derivative of F(x) is the graph such that, for any x, the value of x on the graph of the derivative of F(x) is the slope at point x in F(x).
Jerk is the derivative of acceleration.
a derivative of snip
It means that the first derivative is a constant. The derivative may be with regard to time or any other variable.