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.

Answered by Francesca K. Maths tutor

3003 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given that y = exp(2x) * (x^2 +1)^(5/2), what is dy/dx when x is 0?


f(x) = (4x + 1)/(x - 2) with x > 2. Find a value for 'x' such that f'(x) (first derivative of f(x) with respect to x) is equal to -1.


g(x) = e^(x-1) + x - 6 Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, where x<6


How to differentiate tan(x)?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences