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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

is a constant so it is a multiplier of the derivative
of r. The rule that applies is that the derivative of

is

In this first term the exponent of r is 1. So you are taking the derivative of

which
the rule tells you in

and this simplifies to

.
Putting this all together for the first term, the derivative is

.
.
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

to an equivalent form

.
.
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

.
Again we use the rule that the derivative of

is

.
So the
derivative of

is

which simplifies to

.
Don't forget that this gets multiplied by the constant (2v) so that the derivative
for this second term is

. Multiplying this out results in:
.

and since a negative exponent in the numerator becomes a positive
exponent in the denominator, we could also write the derivative as:
.

.
.
Finally we combine the derivatives for the first and second terms to get:
.

.
and taking care of the signs for the second term concludes the effort by producing
the result:
.

.
Hope this helps you to see your way through the problem.