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