Question

Find the second derivative at $x= 0$ i.e., \[{{\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=0}}\] of the following equation:
 \[{{e}^{y}}+xy=e\]

Answer 1

Hint: Differentiate the given function twice using sum and product rule of differentiation and then evaluate the value of the second derivative at the given point.

Complete step-by-step answer:
We have the function of the form \[{{e}^{y}}+xy=e\].
Differentiating the given function with respect to \[x\] on both sides, we get \[\dfrac{d}{dx}\left({{e}^{y}}+xy \right)=\dfrac{d}{dx}(e)\].
We will use sum rule of differentiation of two functions which states that if \[y=f\left( x
\right)+g\left( x \right)\] then we have \[\dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x
\right)+\dfrac{d}{dx}g\left( x \right)\].

Thus, we have \[\dfrac{d}{dx}\left( {{e}^{y}}+xy \right)=\dfrac{d}{dx}\left( {{e}^{y}}
\right)+\dfrac{d}{dx}\left( xy \right)=\dfrac{d}{dx}\left( e \right).....\left( 1 \right)\] .
We know that the derivative of a constant with respect to any variable is zero.
Thus, we have \[\dfrac{d}{dx}(e)=0.....\left( 2 \right)\].
To evaluate the value of \[\dfrac{d}{dx}\left( {{e}^{y}} \right)\], multiply and divide the equation by \[dy\].
Thus, we have \[\dfrac{d}{dx}\left( {{e}^{y}} \right)=\dfrac{d}{dy}\left( {{e}^{y}} \right)\times
\dfrac{dy}{dx}\].
We know that derivative of function of the form \[y={{e}^{x}}\] is \[\dfrac{dy}{dx}={{e}^{x}}\].
So, we have \[\dfrac{d}{dx}\left( {{e}^{y}} \right)=\dfrac{d}{dy}\left( {{e}^{y}} \right)\times
\dfrac{dy}{dx}={{e}^{y}}\dfrac{dy}{dx}.....\left( 3 \right)\].
We know that the product rule of differentiation says that if \[y=f\left( x \right)g\left( x \right)\] then
we have \[\dfrac{dy}{dx}=f\left( x \right)\dfrac{d}{dx}g\left( x \right)+g\left( x
\right)\dfrac{d}{dx}f\left( x \right)\].
Thus, we have \[\dfrac{d}{dx}\left( xy \right)=x\dfrac{dy}{dx}+y\dfrac{d}{dx}\left( x \right).....\left( 4\right)\].
We know that the derivative of function of the form \[y=a{{x}^{n}}\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
So, we have \[\dfrac{d}{dx}\left( x \right)=1.....\left( 5 \right)\].
Substituting equation \[\left( 5 \right)\] in equation \[\left( 4 \right)\], we get \[\dfrac{d}{dx}\left( xy\right)=x\dfrac{dy}{dx}+y\dfrac{d}{dx}\left( x \right)=x\dfrac{dy}{dx}+y.....\left( 6 \right)\].
Substituting equation \[\left( 2 \right),\left( 3 \right),\left( 6 \right)\] in equation \[\left( 1 \right)\],
we get \[{{e}^{y}}\cdot \dfrac{dy}{dx}+y+x\cdot \dfrac{dy}{dx}=0\].
Simplifying the above equation, we get \[\dfrac{dy}{dx}=-\dfrac{y}{{{e}^{y}}+x}.....\left( 7 \right)\].
We will now differentiate the above equation to find the second derivative of the given function.
As we know \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\].
Thus, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -\dfrac{y}{{{e}^{y}}+x} \right)\].
To find the above derivative, we will use quotient rule which says that if \[y=\dfrac{f\left( x
\right)}{g\left( x \right)}\] then \[\dfrac{dy}{dx}=\dfrac{g\left( x \right)f'\left( x \right)-f\left( x\right)g'\left( x \right)}{{{g}^{2}}\left( x \right)}\].
Thus, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -\dfrac{y}{{{e}^{y}}+x}
\right)=\dfrac{\left( {{e}^{y}}+x \right)\dfrac{d}{dx}\left( -y \right)-\left( -y \right)\dfrac{d}{dx}\left({{e}^{y}}+x \right)}{{{\left( {{e}^{y}}+x \right)}^{2}}}.....\left( 8 \right)\].

To evaluate the value of \[\dfrac{d}{dx}\left( -y \right)\], we will multiply and divide the equation by \[dy\].
Thus, we have \[\dfrac{d}{dx}\left( -y \right)=\dfrac{d}{dy}\left( -y \right)\times \dfrac{dy}{dx}\].
We have \[\dfrac{d}{dy}\left( -y \right)=-1\] as we know that the derivative of function of the form \[y=a{{x}^{n}}\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d}{dx}\left( -y \right)=\dfrac{d}{dy}\left( -y \right)\times \dfrac{dy}{dx}=-
\dfrac{dy}{dx}.....\left( 9 \right)\].
To evaluate the value of \[\dfrac{d}{dx}\left( {{e}^{y}}+x \right)\], we will use sum rule of
differentiation.
Thus, we have \[\dfrac{d}{dx}\left( {{e}^{y}}+x \right)=\dfrac{d}{dx}\left( {{e}^{y}}
\right)+\dfrac{d}{dx}\left( x \right)\]
Substituting equation \[\left( 3 \right),\left( 5 \right)\] in the above equation, we have
\[\dfrac{d}{dx}\left( {{e}^{y}}+x \right)=\dfrac{d}{dx}\left( {{e}^{y}} \right)+\dfrac{d}{dx}\left( x
\right)={{e}^{y}}\dfrac{dy}{dx}+1.....\left( 10 \right)\].
Substituting equation \[\left( 9 \right),\left( 10 \right)\] in equation \[\left( 8 \right)\], we get
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{e}^{y}}+x \right)\dfrac{d}{dx}\left( -y \right)-\left( -y
\right)\dfrac{d}{dx}\left( {{e}^{y}}+x \right)}{{{\left( {{e}^{y}}+x \right)}^{2}}}=\dfrac{\left( {{e}^{y}}+x
\right)\left( -\dfrac{dy}{dx} \right)+y\left( 1+{{e}^{y}}\dfrac{dy}{dx} \right)}{{{\left( {{e}^{y}}+x
\right)}^{2}}}.....\left( 11 \right)\] .
Now, we will evaluate the values of functions at \[x=0\].
Substituting the value \[x=0\] in the equation \[{{e}^{y}}+xy=e\], we have \[{{e}^{y}}=e\].
\[\Rightarrow y=1\]
Substituting the value \[x=0\] in the equation \[\left( 7 \right)\] ,i.e., \[\dfrac{dy}{dx}=-
\dfrac{y}{{{e}^{y}}+x}\], we have \[\dfrac{dy}{dx}=\dfrac{-1}{e}\].
Substituting the value \[x=0\] and \[\dfrac{dy}{dx}=\dfrac{-1}{e}\] in the equation \[\left( 11
\right)\],i.e.,\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{e}^{y}}+x \right)\left( -\dfrac{dy}{dx}
\right)+y\left( 1+{{e}^{y}}\dfrac{dy}{dx} \right)}{{{\left( {{e}^{y}}+x \right)}^{2}}}\], we get
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{e\left( \dfrac{1}{e} \right)+\left( 1+e\left( \dfrac{-1}{e} \right)
\right)}{{{\left( e \right)}^{2}}}\]
Thus, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{e}^{2}}}\] at \[x=0\].

Note: We need to observe that we can’t substitute the values and then find the derivative. We need to find the derivative first and then substitute the values. This is an implicit function which is expressed in terms of both dependent and independent variables. To differentiate an implicit function, we differentiate both sides of an equation by treating one of the variables as a function of the other one.