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

3603 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Integrate (3x^2+2x^-1) with respect to x in the range of K to 3 and explain why K cannot be 0


Find the value of (cos(x) + sec(x))^2 with respect to x when evauated between pi/4 and 0


When you are working out dy/dx = 0, why do you do this and what does it mean?


Two fair six sided dice, called A and B, are rolled and the results are added together. The sum of the dice is 8, what is the probability that two fours were rolled?


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