10 Points- Best answer; Easy Calculus Theory Question.?

Question

Basically, I'm having a momentary elapse, and need to know, why:

when y = k (k is a constant, e.g 1, 4, 2849438918503..e.c.t)

when differentiate it,

why is it that y' = 0.

Please explain it to me in words. Because if you pull down the power, its to the power of 0, and that should just turn it into a 1 rite xD?

Ah, kk sweet,thanks guys, I'll rate the best answer when I log onto my comp next. Thanks everybody!

Answer 1

When differentiating all constant terms go to 0. It is because it doesn't have a variable with it that it goes to 0.

its like this

y=K(x)^0
or y=K
K is constant meaning not changing. Derivatives find the change in the variable thus if all you have is a constant there is no change and the derivative is 0.

Answer 2

You need to remind yourself that y is not just a number. y is a function (e.g. of x). Function

y(x)=k

means, for any x, the function y(x) is always k, and never changes. You can think of derivative as the rate at which y value changes with x. Clearly, y never changes, so the rate is 0, or, equivalently, y'=0.

Answer 3

The first derivative is the rate of change. A constant doesn't have a rate of change, so the result is zero.

Answer 4

By definition :-
f `(x) = lim h->0 [ f (x + h) - f(x) ] / h
In this case , f (x + h) = k = f (x)
f `(x) = lim h->0 [ 0 - 0 ] / h
f `(x) = 0

Answer 5

no, it's technically k * x^0, and when you differentiate, you get 0 * (k / x).