If \[pv = 81\] then \[\dfrac{{dp}}{{dv}}\] at \[v = 9\] is equal to:
A. \[1\]
B. \[ - 1\]
C. \[2\]
D. None of these

Hint: In this question, we need to find the value of \[\dfrac{{dp}}{{dv}}\] at \[v = 9\]. For this, we need to differentiate \[p\] with respect to \[v\]. After that, putting \[v = 9\] in the expression for \[\dfrac{{dp}}{{dv}}\], we get the desired result.

Formula used: The rule for the derivative of \[\dfrac{{f\left( x \right)}}{{g\left( x \right)}}\] is given by
\[\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) - f\left( x \right)\dfrac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\]
Here, \[f\left( x \right)\] and \[g\left( x \right)\] are any two functions of \[x\].

Complete step-by-step answer:
We know that \[pv = 81\]
Thus, we get
\[p = \dfrac{{81}}{v}\]
Now, differentiate the above equation with respect to \[v\].
Thus, we get
\[\dfrac{{dp}}{{dv}} = \dfrac{d}{{dv}}\left( {\dfrac{{81}}{v}} \right)\]
But we know that \[\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) - f\left( x \right)\dfrac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\]
\[\dfrac{{dp}}{{dv}} = \dfrac{{v\dfrac{d}{{dv}}\left( {81} \right) - 81\dfrac{d}{{dv}}\left( v \right)}}{{{v^2}}}\]
By simplifying, we get
\[\dfrac{{dp}}{{dv}} = \dfrac{{v\left( 0 \right) - 81\left( 1 \right)}}{{{v^2}}}\]
\[\dfrac{{dp}}{{dv}} = \dfrac{{ - 81}}{{{v^2}}}\]
Now, put \[v = 9\] in the above equation.
\[\dfrac{{dp}}{{dv}} = \dfrac{{ - 81}}{{{{\left( 9 \right)}^2}}}\]
\[\dfrac{{dp}}{{dv}} = \dfrac{{ - 81}}{{81}}\]
By simplifying further, we get
\[\dfrac{{dp}}{{dv}} = - 1\]
Hence, the value of \[\dfrac{{dp}}{{dv}}\]at \[v = 9\] is \[ - 1\].
Therefore, the correct option is (B).

Additional Information: In mathematics, the derivative is defined as the rate of change of a function with regards to a variable. Derivatives are essential for solving calculus and differential equation problems. Here we can say that the process of determining the derivative is termed differentiation. The derivative is often used to calculate the response of one variable (the dependent variable) to some other variable (independent variable). The first order derivatives suggest if the function is increasing or decreasing in magnitude. The first derivative, also known as the first-order derivative, can be thought of as instantaneous changes that occur. The gradient of the tangent line can be used to predict it.

Note: Here, students generally make mistakes in finding the derivative. To get the correct result of the derivative it is necessary to use the proper rule here. The required answer totally depends on the expression that is obtained after differentiation.