calc quick question?

Question

what is the nth derivative of y=e^nx

would it be ne^(n(x-1))?

Answer 1

y=e^nx
Let's figure out the first three derivatives of it.
y=e^nx
y' = ne^nx
y'' = n²e^nx
y''' = n³e^nx

So there appears to be a pattern. The nth derivative would be (n^n)(e^nx)

Answer 2

Let's find a pattern.

y = e^(nx)
y' = ne^(nx)
y'' = (n^2)e^(nx)
y''' = (n^3)e^(nx)

It appears that for the nth derivative,

y(n) = (n^n)e^(nx)


As a side note, the derivative of y = e^x is NOT xe^(x - 1) because the power rule only works if the base is a variable and the power is a constant.

Hope that helps.

Answer 3

what is the nth derivative of y=e^nx
the 1st derivative is ne^nx
the 2nd derivative would be n^2e^nx
the 3rd derivative would be n^3e^nx
the nth derivative would be n^ne^nx

Answer 4

No, the exponent of e won't change. So the nth derivative is:

y(n) = (n^n)e^(nx)

Answer 5

hi the common fee of a function, f(x), over the period [a, b] might want to be got here upon by: a million/(b - a) * (a to b)? f(x) dx right here, we've a = -7, b = 7, and f(x) = 2x + 14, so we get: a million/(b - a) * (a to b)? f(x) dx = a million/(7 + 7) * (-7 to 7)? 2x + 14 dx = a million/14 * (-7 to 7)? 2x + 14 dx = a million/14 * [x^2 + 14x] | (-7 to 7) = a million/14 * {[(7)^2 + 14(7)] - [(-7)^2 + 14(-7)]} = a million/14 * 196 = 14 So the common fee of the function f(x) = 2x + 14 on the period [-7, 7] is 14. i wish this allows!

Answer 6

a quick question eserves a quick answer : no.


e^(nx) ' = ne^(nx)

ne^(nx)' = n^2e^(nx)

etc

thus the k_the derivative

n^ke^(nx)

you are mixxing another rule : px^n' = pnx^(n-1) but that doesnt apply here

Answer 7

y'=n * e^(nx)

The power does not change.