Differentiate x^3(sinx) with respect to x

As we are differentiating a product (two things times together) we can use the product rule which is if:

                       y = u(x)v(x)

then

                  dy/dx = u(dv/dx) + v(du/dx).

So firstly looking at our equation we need to identify u(x) and v(x). In our case

u(x) = x3 ​        and       v(x) = sinx

Now we need to differentiate both of them seperatly so (remember when we differentiate we times by the old power and then subtract a power)

du/dx = 3x​2          ​and       dv/dx = cosx

Now putting all this into the formula we have

    dy/dx = u(dv/dx) + v(du/dx)

             = x3​cosx + sinx(3x2​)

Then rearranging this we get

        dy/dx = x​3​cosx + 3x2sinx

SC
Answered by Sophie C. Maths tutor

32050 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the derivative of f(x)=x^3 sin(x)


Solving 2tan(x) - 3sin(x) = 0 for -pi ≤ x < pi


What are the necessary conditions for a random variable to have a binomial distribution?


The curve C has the equation 4x^2 - y^3 - 4xy + 2y = 0 . The point P with coordinates (-2, 4) lies on C. Find the exact value of dy/dx at the point P.


We're here to help

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

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning