Suppose that f(x) and k(x) are differentiable everywhere and that f(3) = 5, f'(3) = 7, k(3) = 6, and k'(3) = 9. Find an equation of a tangent line to 3(f(x))^2 at x = 3.

Question

So the derivative of 3(f(x))^2 is 6f(x)*f'(x), and 6f(3)*f'(3) is 210. But for the tangent line, should I use f(3) = 5 for the y-value, or should I use 3(f(3))^2 = 75 instead?

Answer 1

the slope of 3f^2 is 6f f' = 6*5*7=210
So far, so good.

3f^2(3) = 3*5^2 = 75
So, you want the line through (3,75) with slope=210

y-75 = 210(x-3)

your surmise is correct