How to differentiate y=x^3+4x+1 when x=3

First of all you calculate dy/dx. To do this you look at each x individually. For the first x you multiply it by it's power and then minus 1 from the power to get 3x^2. Then multiply the next x by the power and minus 1 to get 4x. As 1 is the same as 1 multiplied by x^0, this means that you multiply 1 by 0 to get 0. Therefore dy/dx is 3x^2 + 4. Now substitute in x=3 to get dy/dx=31. This is the gradient.

LS
Answered by Leana S. Maths tutor

2989 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the gradient of the curve (x^3)-4(y^2)=12xy at the point P(-8,8)


Find the stationary points on the curve y = x^3 + 3x^2 - 9x - 4


A curve is defined by the parametric equations: X = 3 – 4t , y = 1 + (2/t) Find (dy/dx) in terms of t.


Differentiate with respect to x: i) y=x^3ln(2x) ii) y=(x+sin(2x))^3


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning