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

8373 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A curve has equation y = f(x) and passes through the point (4, 22). Given that f ′(x) = 3x^2 – 3x^(1/2) – 7, use integration to find f(x), giving each term in its simplest form.


The curve C has the equation y=((x^2+4)(x-3))/2*x where x is not equal to 0 . Find the tangent to the curve C at the point where x=-1 in the form y=mx+c


If (m+8)(x^2)+m=7-8x has two real roots show that (m+9)(m-8)<0 where m is an arbitrary constant


Find the derivative of the following function with respect to x. y = 5e^x−2xsin(x)


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