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.

Related Maths A Level answers

All answers ▸

Find the sum and product of the roots of the equation 2x^2+3x-5=0


A curve has equations: x=2sin(t) and y=1-cos(2t). Find dy/dx at the point where t=pi/6


When do you know to use integration by parts?


Differentiate x^3(sinx) with respect to x