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

3257 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I integrate sin^2(x)?


Find the integral of xcosx(dx)


A particle of mass 0.25 kg is moving with velocity (3i + 7j) m s–1, when it receives the impulse (5i – 3j) N s. Find the speed of the particle immediately after the impulse.


Given that y= x^(-3/2) + (1/2)x^4 + 2, Find: (a) the integral of y (b) the second differential of y


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