Question

Solve the following that : \[\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}}\]. Given that : $u = \dfrac{{x + y}}{{x - y}}$.
(a) $\dfrac{2}{{{{\left( {x - y} \right)}^2}}}$
(b) Cannot be determined
(c) $\dfrac{2}{{x - y}}$
(d) None of these

Answer 1

Hint: The given problem revolves around the concept's derivatives. To find the solution we will derive the given equation with both ‘$x$’ as well as ‘$y$’ variables. After successfully derivating the terms using laws of derivations i.e. dividation then substituting it in the required solution to get the desired outcome.

Complete step-by-step answer:
Since, we have given the equation that
$u = \dfrac{{x + y}}{{x - y}}$
So, let us assume that ‘$f\left( u \right)$’ as the function of the given equation, we get
\[f\left( u \right) = \dfrac{{x + y}}{{x - y}}\]
First of all, derivating the given function with respect to ‘$x$’, we get
Hence, ‘$y$’ is constant !
(Using law/s of derivative for dividation that is$\dfrac{d}{{dx}}\left( {\dfrac{{numerator}}{{{\text{denominator}}}}} \right) = \dfrac{{\left( {{\text{denominator}}} \right)\dfrac{d}{{dx}}\left( {numerator} \right) - \left( {numerator} \right)\dfrac{d}{{dx}}\left( {{\text{denominator}}} \right)}}{{{{\left( {{\text{denominator}}} \right)}^2}}}$
\[f'\left( u \right) = \dfrac{{\partial u}}{{\partial x}} = \dfrac{{\left( {x - y} \right)\dfrac{d}{{dx}}\left( {x + y} \right) - \left( {x + y} \right)\dfrac{d}{{dx}}\left( {x - y} \right)}}{{{{\left( {x - y} \right)}^2}}}\]
Solving the equation predominantly, we get
\[f'\left( u \right) = \dfrac{{\left( {x - y} \right)\left( {1 + 0} \right) - \left( {x + y} \right)\left( {1 - 0} \right)}}{{{{\left( {x - y} \right)}^2}}}\]
Solving the equation mathematically, we get
\[f'\left( u \right) = \dfrac{{\left( {x - y} \right) - \left( {x + y} \right)}}{{{{\left( {x - y} \right)}^2}}}\]
\[f'\left( u \right) = \dfrac{{x - y - x - y}}{{{{\left( {x - y} \right)}^2}}}\]
Hence, adding and subtracting the terms, we get
\[f'\left( u \right) = \dfrac{{ - 2y}}{{{{\left( {x - y} \right)}^2}}}\] … ($1$)
Similarly,
Hence, ‘$x$’ is constant !
Derivating the given function with respect to ‘$y$’, we get
(Using law/s of derivative for dividation that is$\dfrac{d}{{dx}}\left( {\dfrac{{numerator}}{{{\text{denominator}}}}} \right) = \dfrac{{\left( {{\text{denominator}}} \right)\dfrac{d}{{dx}}\left( {numerator} \right) - \left( {numerator} \right)\dfrac{d}{{dx}}\left( {{\text{denominator}}} \right)}}{{{{\left( {{\text{denominator}}} \right)}^2}}}$
\[f'\left( u \right) = \dfrac{{\partial u}}{{\partial y}} = \dfrac{{\left( {x - y} \right)\dfrac{d}{{dx}}\left( {x + y} \right) - \left( {x + y} \right)\dfrac{d}{{dx}}\left( {x - y} \right)}}{{{{\left( {x - y} \right)}^2}}}\]
Solving the equation predominantly, we get
\[f'\left( u \right) = \dfrac{{\left( {x - y} \right)\left( {0 + 1} \right) - \left( {x + y} \right)\left( {0 - 1} \right)}}{{{{\left( {x - y} \right)}^2}}}\]
Solving the equation mathematically, we get
\[f'\left( u \right) = \dfrac{{\left( {x - y} \right) + \left( {x + y} \right)}}{{{{\left( {x - y} \right)}^2}}}\]
\[f'\left( u \right) = \dfrac{{x - y + x + y}}{{{{\left( {x - y} \right)}^2}}}\]
Hence, adding and subtracting the terms, we get
\[f'\left( u \right) = \dfrac{{2x}}{{{{\left( {x - y} \right)}^2}}}\] … ($2$)
Now, as a result from ($1$) and ($2$),
The final solution for the problem is about to solve since by substituting the values that we have solved,
\[\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}} = \dfrac{{ - 2y}}{{{{\left( {x - y} \right)}^2}}} + \dfrac{{2x}}{{{{\left( {x - y} \right)}^2}}}\]
Solving the equation mathematically, we get
\[\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}} = \dfrac{{2x - 2y}}{{{{\left( {x - y} \right)}^2}}}\]
\[\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}} = \dfrac{{2\left( {x - y} \right)}}{{{{\left( {x - y} \right)}^2}}}\]
As a result, multiplying the equation by\[\left( {x - y} \right)\], we get
\[\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}} = \dfrac{2}{{x - y}}\]
$\therefore \Rightarrow $The option (c) is correct!
So, the correct answer is “Option c”.

Note: One must know the laws of the derivatives such as multiplication, dividation that is$\dfrac{d}{{dx}}\left( {\dfrac{{numerator}}{{{\text{denominator}}}}} \right) = \dfrac{{\left( {{\text{denominator}}} \right)\dfrac{d}{{dx}}\left( {numerator} \right) - \left( {numerator} \right)\dfrac{d}{{dx}}\left( {{\text{denominator}}} \right)}}{{{{\left( {{\text{denominator}}} \right)}^2}}}$, etc. Also, sure that derivation of function$f(x)$ is also represented as\[f'(x) = \dfrac{{dy}}{{dx}} = \dfrac{{\partial y}}{{\partial x}}\] and so on. Absolute implementation of solving should be done so as to be sure of our final answer.