I too am confused. Traditionally the constant c arises due to integration, if so the equation would become:Papabravo said:I'm confused, c does not appear in the function, or am I missing something?
d/dx(ln(x)) / d/dx(x-1) = (1/x) /1 = 1/x evaluated at x = 1 is just 1
To correct the above:Papabravo said:To evaluate the limit of an indeterminate form use L'hopital's rule. Take the derivative of the numerator and the derivative of denominator. Now evaluate that quotient.
so c = 1 is the value of the function which makes the original function continuous.Rich (BB code):
d/dx(ln(x)) / d/dx(x-1) = (1/x) /x = 1/x^2 evaluated at x = 1 is just 1
|Thread starter||Similar threads||Forum||Replies||Date|
|Calculus Chain Rule Problem. Any help would be greatly appreciated.||Homework Help||2|
|P||Question about "Inductors and Calculus"||Analog & Mixed-Signal Design||19|
|What would it take to learn Calculus at a practical level?||General Science, Physics & Math||55|
|O||Calculus. Derivatives of inverse trig functions||General Science, Physics & Math||4|
|H||Calculus problem I have not seen before||Homework Help||3|
by Steve Arar
by Lianne Frith
by Kate Smith