Differentiating (x^2)(sinx) Using the Product Rule

Firstly, what is the product rule? What does it actually say? Well, it tells us how to differentiate a function of the form uv - the product of the functions u and v. If y = uv, the product rule says:

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

So you differentiate the first bit, leaving the second part alone - giving you (du/dx)v. Then, you differentiate the second bit, leaving the first alone, - giving you u(dv/dx). And then you just add the two results to get dy/dx.

Let's look at applying this to the example in the question, trying to differentiate this: y = (x^2)(sinx)

We can see that y is a product of two functions, x^2 and sinx. Using the process above, we differentiate the first part, x^2, and leave sinx alone. That gives us (2x)(sinx). Then, we differentiate the second bit, sinx, and leave x^2 alone, and that gives us (x^2)(cosx). Then we just add the two together: (2x)(sinx) + (x^2)(cosx). So from that calculation we've shown that

dy/dx = (2x)(sinx) + (x^2)(cosx)

EM
Answered by Edward M. Maths tutor

5061 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Rationalise the fraction : 5/(3-sqrt(2))


If I throw a ball vertically upwards with a velocity of 15 m/s and we assume the gravitational acceleration is 10 m/s^2. Draw the distance-time, and velocity-time graphs, how long is the ball in the air before it returns to the point I threw it from?


If y = exp(x^2), find dy/dx


Express 4sinx + 3cosx in the form Rcos(x-a)


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:

© 2025 by IXL Learning