Given that y = ((4x+3)^5)(sin2x), find dy/dx

First of all, we have to use the product rule, since two things are multiplied together.  The product rule states that d/dx (u*v) = vu' + uv'

Let u = (4x+3)5  v = sin(2x)

Now, to find  u' and v' we have to find du/dx and dv/dx. As we see, we need to use the chain rule to find du/dx

u' = 20(4x+3)4   v' = 2cos(2x)

Finally, dy/dx = vu' + uv' = 20sin(2x)(4x+3)4 + 2cos(2x)(4x+3)5

JN
Answered by Juozas N. Maths tutor

2865 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Solve for x (where 0<x<360) 2sin^2(x) - sin(x) - 1 = 0


Express the fraction (p+q)/(p-q) in the form m+n√2, where p=3-2√2 and q=2-√2.


Show that the curve with equation y=x^2-6x+9 and the line with equation y=-x do not intersect.


Write 9sin(x) + 12 cos(x) in the form Rsin(x+y) and hence solve 9sin(x) + 12 cos(x) = 3


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences