Question

If \[pv{\text{ }} = {\text{ }}81\] , then\[\dfrac{{dp}}{{dv}}\] at \[v{\text{ }} = {\text{ }}9\] is equal to
\[\left( 1 \right)\] \[1\]
\[\left( 2 \right)\] \[ - 1\]
\[\left( 3 \right)\] \[2\]
\[\left( 4 \right)\] \[none{\text{ }}of{\text{ }}these\]

Answer 1

Hint: We have to find the derivative of \[p\] with respect to $v$. Using quotient rule of derivative $d[\dfrac{u}{v}] = \dfrac{{(u \times v' - v \times u')}}{{{v^2}}}$provided all functions are defined . Make the equation in terms of a single variable \[p\] , then differentiate it with respect to \[v\] gives \[\dfrac{{dp}}{{dv}}\]. After applying the quotient rule we will put the value of \[v\]. This gives the value of \[{\text{ }}\dfrac{{dp}}{{dv}}\].

Complete step-by-step answer:
Differentiation, in mathematics , is the process of finding the derivative , or the rate of change of a given function . In contrast to the abstract nature of the theory behind it , the practical technique of differentiation can be carried out by purely algebraic manipulations , using three basic derivatives , four rules of operation , and a knowledge of how to manipulate functions. We can solve any of the problems using the rules of operations i.e. addition , subtraction , multiplication and division .
Given : \[pv{\text{ }} = {\text{ }}81\]
Simplifying , \[p{\text{ }} = \dfrac{{81}}{v}\] ——(1)
Differentiate \[p\] with respect to and applying quotient rule , we get
$\dfrac{{dp}}{{dv}} = \dfrac{{81[0 \times v - 1 \times 1]}}{{{v^2}}}$
$\dfrac{{dp}}{{dv}} = 81[\dfrac{{ - 1}}{{{v^2}}}]$ ——(2)
We have to find value of \[\dfrac{{dp}}{{dv}}\] at \[v{\text{ }} = {\text{ }}9\]
Put \[v{\text{ }} = {\text{ }}9\] in (2) , we get
$\dfrac{{dp}}{{dv}} = \dfrac{{ - 81}}{{{{(9)}^2}}}$
\[\dfrac{{dp}}{{dv}}{\text{ }} = {\text{ }} - 1\]
Thus , the correct option is \[\left( 2 \right)\]
So, the correct answer is “Option 2”.

Note: Derivative of sum of two function is equal to sum of the derivatives of the functions :
\[\dfrac{{d\left[ {f\left( x \right){\text{ }} + {\text{ }}g\left( x \right){\text{ }}} \right]}}{{dx}}{\text{ }} = {\text{ }}\dfrac{{d{\text{ }}f\left( x \right)}}{{dx}}{\text{ }} + \dfrac{{d{\text{ }}g\left( x \right)}}{{dx}}\]
Derivative of product of two function is difference of the derivatives of the functions :
\[{\text{ }}\dfrac{{d\left[ {f\left( x \right){\text{ }} - {\text{ }}g\left( x \right)} \right]}}{{dx}} = {\text{ }}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} - {\text{ }}\dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}\]
Derivative of product of two function is given by the following product rule :
\[\dfrac{{d\left[ {f\left( x \right){\text{ }} \times {\text{ }}g\left( x \right)} \right]}}{{dx}}\] \[ = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} \times g{\text{ }} + {\text{ }}f{\text{ }} \times {\text{ }}\dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}\]
Derivative of quotient of two function is given by the following quotient rule :
\[\dfrac{{d\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right]}}{{dx}}= \dfrac{{[\dfrac{{d[f(x)]}}{{dx}} \times g(x) - f(x) \times \dfrac{{d[g(x)]}}{{dx}}]}}{{{{[g(x)]}^2}}}\]