Find the gradient of 4(8x+2)^4 at X coordinate 2

To find a gradient at a given point, first we differentiate then we sub in the x coordinate of the point. To differentiate 4(8x+2)4 we must use the chain rule. First we let u= 8x+2 and differentiate this to find du/dx = 8. Then we must find the rest, we differentiate y = 4u4 , dy/du = 16u3 . The chain rule then states that dy/dx = dy/du x du/dx so we get 16u3 x 8 and as u = 8x+2 we get: 128(8x+2)3.To find the value at X coordinate 2 we sub the value into dy/dx and get 128(8(2)+2)3 multiply this out and we get 128 x 183 = 746,946. So the gradient of 4(8x+2)4 at X coordinate 2 is 746,946.

Answered by Maths tutor

3874 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given that y = 4x^3 – 5/(x^2) , x =/= 0, find in its simplest form dy/dx.


Given that: y = 3x^2 + 6x^1/3 + (2x^3 - 7)/(3x^1/2), x > 0 Find dy/dx, give each term in its simplest form


Differentiate y = x^3 + 2x^2 + 4x + 3


Solve for x, 5sin(x) - 3cos(x) = 2 , in the interval 0<x<2pi


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:

© 2026 by IXL Learning