Question

If \[{e^y} + xy = e\], then find the value of \[\dfrac{{{d^2}x}}{{d{y^2}}}\] for \[x = 0\].
A. \[\dfrac{1}{e}\]
B. \[\dfrac{1}{{{e^2}}}\]
C. \[\dfrac{1}{{{e^3}}}\]
D. None of these

Answer 1

Hint: In the question, the given equation is an exponential equation. First, we will substitute \[x = 0\] in the given equation and find the value of \[y\]. Then, we will find the first derivative of the given equation with respect to \[x\]. By differentiating the first order derivative of the given equation with respect to \[x\], we will find the value of \[\dfrac{{{d^2}x}}{{d{y^2}}}\]. Then we will put the value of \[x\] and \[y\].
Formula Used:
Product formula: \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]
\[\dfrac{d}{{dx}}\left( {{e^y}} \right) = {e^y}\dfrac{{dy}}{{dx}}\]
Complete step by step solution:
The given equation is \[{e^y} + xy = e\].
Substitute \[x = 0\] in the above equation.
\[{e^y} + \left( 0 \right)y = e\]
\[ \Rightarrow \]\[{e^y} = e\]
\[ \Rightarrow \]\[y = 1\]

Let’s differentiate the given equation with respect to \[x\].
We get,
\[{e^y}\dfrac{{dy}}{{dx}} + x\dfrac{{dy}}{{dx}} + y = 0\] ….(1) [since \[\dfrac{d}{{dx}}\left( e \right) = 0\] as \[e\] is a constant]
Substitute \[x = 0\] and \[y = 1\] in above equation.
\[{e^1}\dfrac{{dy}}{{dx}} + 0\dfrac{{dy}}{{dx}} + 1 = 0\]
\[ \Rightarrow \]\[e\dfrac{{dy}}{{dx}} + 1 = 0\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{e}\]

Differentiate equation (1) with respect to \[x\].
We get,
\[{e^y} \cdot \dfrac{{dy}}{{dx}}\dfrac{{dy}}{{dx}} + {e^y} \cdot \dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} + x \cdot \dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} = 0\]
\[ \Rightarrow \]\[{e^y} \cdot {\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \left( {{e^y} + x} \right) \cdot \dfrac{{{d^2}y}}{{d{x^2}}} + 2\dfrac{{dy}}{{dx}} = 0\]
\[ \Rightarrow \]\[\left( {{e^y} + x} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = - {e^y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - 2\dfrac{{dy}}{{dx}}\]
Substitute \[x = 0\], \[y = 1\] and \[\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{e}\] in the above equation.
\[\left( {{e^1} + 0} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = - {e^1}{\left( {\dfrac{{ - 1}}{e}} \right)^2} - 2\dfrac{{ - 1}}{e}\]
\[ \Rightarrow \]\[e\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{e}{{{e^2}}} + \dfrac{2}{e}\]
\[ \Rightarrow \]\[e\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{1}{e} + \dfrac{2}{e}\]
\[ \Rightarrow \]\[e\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{2e - e}}{{{e^2}}}\]
\[ \Rightarrow \]\[e\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{e}\]
Divide both sides by \[e\].
\[ \Rightarrow \]\[\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{{{e^2}}}\]

Hence the correct option is B.
Note: Students are often confused with the formula \[\dfrac{d}{{dx}}\left( {{e^y}} \right) = {e^y}\dfrac{{dy}}{{dx}}\] and \[\dfrac{d}{{dx}}\left( {{e^y}} \right) = {e^y}\]. But the correct formula is \[\dfrac{d}{{dx}}\left( {{e^y}} \right) = {e^y}\dfrac{{dy}}{{dx}}\]. Because we also have to calculate the derivative of \[y\]with respect to \[x\].