[cheche_15]

What is the first and seconf derivative of 250x^2/e^3x

[cool_dude]

you must use the quotient rule.

Note: derivative of e^3x = 3e^3x

once you get first derivative than use quotient rule again to get second derivative.

Post your answer once you got it and I'll tell you if it is right.:)

[cheche_15]

I get 500x(e^3x)-250x^2(3e^3x)/(e^3t)^2 for the first derivative is that right?

[s_cianci]

Apply the quotient rule twice over. Recall d/dx[u/v] = (vu' - uv')/v^2. For the first iteration, u = 250x^2 and v = e^3x. Obtain u' by applying the Power Rule and obtain v' by applying the Chain Rule, recalling that an exponential derivative with a natural base is an identity.

[s_cianci]

I get 500x(e^3x)-250x^2(3e^3x)/(e^3t)^2 for the first derivative is that right??

Yes. Now apply the quotient rule again.

[cool_dude]

I get 500x(e^3x)-250x^2(3e^3x)/(e^3t)^2 for the first derivative is that right??

Yes cheche_15 That is correct although you did put a "t" for no reason which I know you meant it as an x. Your first derivative is correct good job. Now for the second derivative use the quotient rule again and you should get the second derivative. Post your answer for the second derivative and we'll tell you if it correct.