Let f(x) = 5x^4 + 6x^3 + 3, find dy/dx at x = 3
First we must differentiate the equation with respect to x. To differentiate you must multiply the coefficient (number in front) by the power of x, then subtract 1 from the power. So here we find dy/dx = (54)x^(4-1) + (63)x^(3-1) + (1*0) = 20x^3 + 18x^2.
To find the value of dy/dx at x=3 we must substitute x=3 into the equation we just found. This gives dy/dx = (203^3) + (183^2) = 540 + 162 = 702. This value is the gradient of the line at x=3.
FK
Related Maths A Level answers
All answers ▸How do I show that (cos^4x - sin^4x) / cos^2x = 1 - tan^2x
A curve C has equation y = x^2 − 2x − 24 x^(1/2), x > 0 (a) Find (i) dy/d x (ii) d^2y/dx^2 (b) Verify that C has a stationary point when x = 4 (c) Determine the nature of this stationary point, giving a reason for your answer.
Calculate the binomial expansion of (2x+6)^5 up to x^3 where x is decreasing.
Using implicit differentiation, write the expression "3y^2 = 4x^3 + x" in terms of "dy/dx"