1. The problem statement, all variables and given/known data Find an equation of the tangent line at the point indicated f(x) = 5x2-2x+9 , x = 1 2. Relevant equations (d/dx) bx = ln(b)bx General Power Rule which states: (d/dx) g(x)n = n(g(x))n-1 * g'(x) 3. The attempt at a solution So looking at a previous problem first, we have s(t) = 37t, t = 2 So, we can substitute u for the exponent, and take the derivative of that. We get (ln 3) * 3u * (7t)' which equals (7 ln 3)314 Since it wants to find the tangent line we have to take it further. The equation to the tangent line is y-f(a) = f'(a)(x-a) and point slope form is y = f'(a)(x-a) + f(a) It helps me to look at it in this scenario as f'(a) = solution before putting it in slope form, a = the given x value(or t in this case), and f(a) = bx Using this first example problem that would mean y = 7 ln 3 * 314(t-2) + 314, and the book confirms this. Now, trying to generalize these principles to the problem I am working on.. I switch the exponent with "u" to simplify the problem. We now have (ln 5)5u(x2-2x+9)' and since x = 1 its (ln 5)5u * 8 which gives us (8 ln 5)58 Attemping to put this into the equation for a tangent line, f'(a) = (8 ln 5)58 a = 1 f(a) = bx = 58 so putting that all together is y = (8 ln 5)58(x-1) + 58.... or so I thought. The book gives y = 58 How did I screw this up?