If y=3x^3e^x; find dy/dx?

Using the product rule we know that dy/dx = uv' + vu' where u = 3x^3; v = e^x. e^x differentiates to itself multiplied by any number in front of the x. u' = 9x^2; v' = e^x. Therefore dy/dx = 3x^3e^x + 9x^2e^x. This could be simplified further if the question asks for the answer in its simplest form. 

AG
Answered by Aimee G. Maths tutor

5240 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given y=2x^4-1+x^1/2, solve dy/dx


Differentiate y = (3x^3+2x+7)/x^(1/2)


Chris claims that, “for any given value of x , the gradient of the curve y=2x^3 +6x^2 - 12x +3 is always greater than the gradient of the curve y=1+60x−6x^2” . Show that Chris is wrong by finding all the values of x for which his claim is not true.


How do i find dy/dx in terms of t for two parametric equations that are in terms of t.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences