Find dy/dx where y= x^3(sin(x))

To differentiate y, we must used the product rule.The product rule is d/dx [f(x)g(x)] = f'(x)g(x) + g'(x)f(x)So here, we let f(x)= x^3 and g(x)= sin(x)Then, f'(x)= 3x^2 and g'(x) = cos(x)Then substituting these into the product rule formula, we get dy/dx = (3x^2)sin(x) + cos(x)x^3We can simplify the answer by factorising out x^2 :dy/dx= x^2[3sin(x) + xcos(x)]

KC
Answered by Kajal C. Maths tutor

8035 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Differentiate with respect to x y=(x^3)ln2x


f(x)= 2x^3 -7x^2 + 2x +3. Given that (x-3) is a factor of f(x), express f(x) in a fully factorised form.


How do you rationalise the denominator?


How do I differentiate a pair of parametric equations?


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