Question

If \[z = \dfrac{y}{x}\left[ {\sin \dfrac{x}{y} + \cos \left( {1 + \dfrac{y}{x}} \right)} \right]\] , then find the value of \[x\dfrac{{\delta z}}{{\delta x}}\] .
A.\[y\dfrac{{\delta z}}{{\delta y}}\]
B. \[ - y\dfrac{{\delta z}}{{\delta y}}\]
C. \[2y\dfrac{{\delta z}}{{\delta y}}\]
D. \[2y\dfrac{{\delta z}}{{\delta x}}\]

Answer 1

Hint: First write the Euler’s formula of partial differential equation then conclude the required result.

Formula used:
Euler’s theorem states that,
If u be a homogeneous function of x, y with degree n, then
\[x\dfrac{{\delta u}}{{\delta x}} + y\dfrac{{\delta u}}{{\delta y}} = nu\].

Complete step by step solution:
The given equation is,
\[z = \dfrac{y}{x}\left[ {\sin \dfrac{x}{y} + \cos \left( {1 + \dfrac{y}{x}} \right)} \right]\]
The degree of this equation is 0 as here x is divided by y or y is divided by x in both the cases the degree of the numerator and the denominator is 1, so if we write the given term as x multiplied by y power -1 then the degree will be 0.
Therefore, by Euler’s theorem we have,
\[x\dfrac{{\delta z}}{{\delta x}} + y\dfrac{{\delta z}}{{\delta y}} = 0 \times z\]
  \[x\dfrac{{\delta z}}{{\delta x}} + y\dfrac{{\delta z}}{{\delta y}} = 0\]
\[x\dfrac{{\delta z}}{{\delta x}} = - y\dfrac{{\delta z}}{{\delta y}}\]
The correct option is B.

Note: Sometimes students don't use Euler's theorem and want to differentiate the given equation directly with respect to x then with respect to y and calculate as required, but that process is lengthy. So, using Euler’s theorem to answer this type of questions it will be easy.