Question 71523
{{{2(pi)(r)-(2v/r^2)}}}
.
The problem asks for you to take the derivative of the above expression. I presume that
the derivative is with respect to r and that v is a constant.
.
One term at a time. The product {{{2(pi)}}} is a constant so it is a multiplier of the derivative
of r.  The rule that applies is that the derivative of {{{r^n}}} is {{{n*r^(n-1)}}}
In this first term the exponent of r is 1. So you are taking the derivative of {{{r^1}}} which
the rule tells you in {{{1*r^(1-1)}}} and this simplifies to {{{1*r^0 = 1*1 = 1}}}.
Putting this all together for the first term, the derivative is {{{2*(pi)*1 = 2*(pi)}}}.
.
On to the second term. The most critical point here is to recognize that a positive
exponent in the denominator is equivalent to a negative exponent in the numerator.
Using this we can convert {{{(2v)/r^2}}} to an equivalent form {{{(2v)*r^(-2)}}}.
.
Now we can apply the same technique as we did for the first term. The factor (2v) is
presumed to be a constant and therefore will be a multiplier of the derivative of {{{r^(-2)}}}.
Again we use the rule that the derivative of {{{r^n}}} is {{{n*r^(n-1)}}}.
So the
derivative of {{{r^(-2)}}} is {{{(-2)*r^(-2-1)}}} which simplifies to {{{(-2)*r^(-3)}}}.
Don't forget that this gets multiplied by the constant (2v) so that the derivative
for this second term is {{{(2v)(-2)r^(-3)}}}. Multiplying this out results in:
.
{{{(-4vr^(-3))}}} and since a negative exponent in the numerator becomes a positive
exponent in the denominator, we could also write the derivative as:
.
{{{-4v/(r^3)}}}.
.
Finally we combine the derivatives for the first and second terms to get:
.
{{{2(pi) - (-4v)/r^3}}}
.
and taking care of the signs for the second term concludes the effort by producing
the result:
.
{{{2(pi) + (4v)/r^3}}}
.
Hope this helps you to see your way through the problem.