Find an equation of the tangent line to the graph of y = ln(x2) at the point (8, ln(64)).

Question

Find an equation of the tangent line to the graph of y = ln(x²) at the point (8, ln(64)).

Give your answer in the slope-intercept form. (Do NOT simplify ln(64) as 2ln(8).)

Answer 1

Hello Mark,

The formula for the equation of the tangent line: f(x) = f'(a) *(x-a) + f(a) with f'(a) being the value of the f'(x) when x =a; a being the x-coordinate of point you're given, and f(a) being its y-coordinate.

Let's find the derivative of f(x): f'(x) = (derivative of x²)/x² = 2x/x² = 2/x

Therefore: f'(a) = f'(8) = 2/8 = 1/4

f(a) = ln 64

You can now set up the equation of the tangent line:

f(x) = (1/4)* (x - 8) + ln(64)

You could also expand the previous answer and get: f(x) = (1/4)*x - 2 + ln(64)