f(x) = KxL
Problem 1: Find f'(x).
"v" and "w" are two points located on the x-axis.
Problem 2: Calculate f'(v).
Problem 3: Find the equation of the tangent line of f(x) over the point "w". The equation should be in this form y=Ax+B
Problem 1:This is how I worked out the differentiation:
f(x) = e(ln 1.78)x-1.39
f'(x) = (Ku)'(u) * u'
(Ku)'(u) = (eu ln K)' = ln K * eu ln K = ln K * Ku
u' = LxL-1 = -1.39x-2.39
f'(x) = ln 1.78(1.78x-1.39) * -1.39x-2.39
Is this correct?
Regarding the other problems, do I need f'(x) to solve these? I don't know where to start.