help please d/dx (e^-ax^2) = ?

Question

d/dx (e^-ax^2) = (-ax^2)' * ( e^-ax^2) ? is it like d/dx (e^ax )= a * e^ax?

Answer 1

d/dx (e^-ax^2) = (-2ax)*e^(-ax^2)



When you take derivative of e^n, you have to take derivative of n power with respect to whatever variable then ask for and then multiply it by original e^n. Hope that helps.

Answer 2

Sometimes you need to apply the chain rule multiple times. Derivative of the outside times the derivative of the inside; keep going until you run out of insides.



d(e^[-ax^2])/dx = e^(-ax^2)*(-2ax)

Answer 3

AS d/dx (e^-ax^2)



=(e^-ax^2) into d/dx (-ax^2)



=(e^-ax^2) into [-a d/dx(x^2)]



=(e^-ax^2) into [-a (2x)]



= -2ax(e^-ax^2)



Regards,

S. Faheem

Answer 4

how about using ln() function... cause thats the inverse of e.

then you use the:

y = ln(x)

dy/dx = f '(x)/f(x)

or you can use the y = e^x then.. f ' (x)e^(f(x))



making it: -2ae^(-ax^2)

Answer 5

         d { eʷ } ⁄ dx = eʷ • (dw ⁄ dx)



w = -a x²

dw ⁄ dx = - 2 a x



         d { eʷ } ⁄ dx = e^(-a x²) • (- 2 a x)



              = (- 2 a x) • e^(-a x²)

Answer 6

y=e^(-ax^2)

dy/dx=-2axe^(-ax^2) answer//

Answer 7

= -2ax*e^(-ax²)

Answer 8

hmmmm....tough question......