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 ▸The curve C has the equation 4x^2 - y^3 - 4xy + 2y = 0 . The point P with coordinates (-2, 4) lies on C. Find the exact value of dy/dx at the point P.
A curve has equation y = f(x) and passes through the point (4, 22). Given that f ′(x) = 3x^2 – 3x^(1/2) – 7, use integration to find f(x), giving each term in its simplest form.
z = 5 - 3i Find z^2 in a form of a + bi, where a and b are real constants
How do you do integration by parts?