Find an equation of the tangent line to the curve at the given point. y = (1 + 4x)12, (0, 1)

THe given equation is

[tex] y=(1+4x)^{12} [/tex]

Differentiating using chain rule

[tex] dy/dx = 4*12(1+4x)^{11} = 48(1+4x)^{11} [/tex]

At point (0,1), we will get

[tex] dy/dx = 48(1+4(0))^{11} = 48 [/tex]

Now we use point slope form, which is

[tex] y-y_{1} = m (x-x_{1}) [/tex]

Substituting the values of m, and x1,y1, we will get

[tex] y-1=48(x-0) [/tex]

[tex] y = 48x+1 [/tex]

And that's the required answer .