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

Related Maths A Level answers

All answers ▸

A curve is defined by the parametric equations x=t^2/2 +1, y=4/t -1. Find the gradient of the curve when t =2.


find the definite integral between limits 1 and 2 of (4x^3+1)/(x^4+x) with respect to x


Given that y = 4x^3 – 5/(x^2) , x not equal to 0, find in their simplest form (a) dy/dx, and (b) integral of y with respect to x.


Differentiate y=ln(x)+5x^2, and give the equation of the tangent at the point x=1