find the derevative (hint try to use limits)
(x^2010 - 1 )\(x -1 )
already answered (separate topic)
Using the L'Hopital limit theorem (at x = 1) will not help you compute the derivative for arbitrary values of x.
The quotient function can be rewritten,
x^2009 + x^2008 + ...+ x^2 + x + 1
I had to leave out 2005 different terms. you then differentiate that and get a long string of terms.
You can also get a shorter expression for the derivative by using the formula for the derivative of the ratio of two functions.
f'(x) = [2010x^2009*(x-1) - x^2010]/(x-1)^2
= 2010*x^2009/(x-1) - x^2010/(x-1)^2
To find the derivative of the function (x^2010 - 1)/(x - 1), we can use the concept of limits and apply the quotient rule. Here's how to find the derivative:
Step 1: Start by applying the quotient rule, which states that the derivative of (u/v) is (u'v - uv') / v^2.
Step 2: Let u = x^2010 - 1 and v = x - 1.
Step 3: Find the derivative of u with respect to x, denoted as u'. In our case, u' = d/dx(x^2010 - 1).
To differentiate u, we need to use the power rule, which states that the derivative of x^n is n*x^(n-1). Applying the power rule, we obtain:
u' = d/dx(x^2010 - 1)
= 2010 * x^(2010-1) - 0 ( derivative of -1 is 0)
= 2010 * x^2009.
Step 4: Find the derivative of v with respect to x, denoted as v'. In our case, v' = d/dx(x - 1).
The derivative of x with respect to x is 1, and the derivative of a constant (in this case, -1) is 0.
v' = d/dx(x - 1)
= 1 - 0
= 1.
Step 5: Apply the quotient rule:
f'(x) = (u'v - uv') / v^2
= [(2010 * x^2009) * (x - 1) - (x^2010 - 1) * 1] / (x - 1)^2
= [2010 * x^(2009+1) - 2010 * x^2009 - x^2010 + 1] / (x - 1)^2
= [2010 * x^2010 - 2010 * x^2009 - x^2010 + 1] / (x - 1)^2.
Therefore, the derivative of (x^2010 - 1)/(x - 1) is [2010 * x^2010 - 2010 * x^2009 - x^2010 + 1] / (x - 1)^2.