Find an equation of the tangent line to the curve at the given point.

Question

Find an equation of the tangent line to the curve at the given point.

y = (1 + 5x)⁸,  (0, 1)

Answer 1

The slope of the tangent line at any point (x,y) is

y' = 8(1+5x)⁷ * 5 = 40(1+5x)⁸

So, at (0,1), y'(1) = 40

Now, using the point-slope form for a line, the equation is

y-1 = 40(x-0)

or,

y = 40x+1

Answer 2

Use the chain rule

f'(x) = 8(1 + 5x)⁷(5)

f'(x) = 40(1 + 5x)⁷

f'(-0.2) = 0

Can you derive the equation?