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 :
$\dfrac{{d\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right]}}{{dx}}= \dfrac{{[\dfrac{{d[f(x)]}}{{dx}} \times g(x) - f(x) \times \dfrac{{d[g(x)]}}{{dx}}]}}{{{{[g(x)]}^2}}}$