Find the equation of the tangent at x=1 for the curve y=(4x^2+1)^3

The tangent is the straight line passing through x=1, touching the curve only at that point. For x=1, y=(4+1)^3=125 Using the chain rule we obtain dy/dx = 38x(4x^2+1)^2. To then get the gradient of the tangent we take dy/dx at x=1. dy/dx[x=1]= 38(4+1)^2 = 600. As tangent is a straight line the equation is in the form y=mx+c with m = dy/dx[x=1]. We simply sub in (x,y)=(1,125) and m=600 to find c. 125=600+c c=-475 So the equation of the tangent is y=600x-475

JH
Answered by Jacob H. Maths tutor

3326 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

If the function f is defined as f= 1-2x^3 find the inverse f^-1


Imagine a sector of a circle called AOB. With center O and radius rcm. The angle AOB is R in radians. The area of the sector is 11cm². Given the perimeter of the sector is 4 time the length of the arc AB. Find r.


A circle with centre C has equation x^2 + y^2 + 2x + 6y - 40 = 0 . Express this equation in the form (x - a)^2 + (x - b)^2 = r^2. Find the co-ordinates of C and the radius of the circle.


Prove by induction that, for n ∈ Z⁺ , [3 , -2 ; 2 , -1]ⁿ = [2n+1 , -2n ; 2n , 1-2n]


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences