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

Related Maths A Level answers

All answers ▸

(Core 3 level) Integrate the function f(x) = 2 -cos(3x) between the bounds 0, pi/3.


How do I draw and sketch an equation?


Integrate 2x^3 -4x +5


Integrate 2sin(theta)cos(2*theta)