Derive.....?

Question

In my current texbook, they don't tell what to do in this instance.
e^-3x

Answer 1

e^x derived is e^x(dx) dx in this case is 1

e^-3x derived is e^-3x(dx) dx n this case is -3
solution is -3e^-3x

Answer 2

The e is subjected to the x, and the -3 is subjected to the x, so use the chain rule.
y = e^-3x
dy/dx = e^-3x(-3)
dy/dx = -3e^-3x

Answer 3

A negative exponent will become a positive exponent of the other side of the fraction, so

e^-3x = 1/e^3x

Hope this helps

Answer 4

Let y = e^(-3x)
let u = - 3x
du/dx = - 3
y = e^u
dy/du = e^u
dy/dx = (dy/du) (du/dx)
dy/dx = e^u (- 3)
dy/dx = (- 3) e^(- 3x)

Answer 5

y=e^-3x
dy/dx=dy/dx(e^-3x )
Let z = -3x
dy/dx=dy/dz(e^z)*dz/dx(z)
=e^z*dz/dx(-3x)
=e^(-3x)*-3
=-3e^-3x