Question

# If${{e}^{x+y}}=xy$, then show that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-y\left( {{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$.

Hint: Directly apply the differentiation to the given expression using the exponential differentiation, product and quotient rule of differentiation. Convert the first order derivative in terms of $'x'$ and $'y'$. Then proceed with finding the second order derivative and simplify it.

The given expression is ${{e}^{x+y}}=xy$
Differentiate the given expression with respect to $'x'$ , we get
$\dfrac{d}{dx}\left( {{e}^{x+y}} \right)=\dfrac{d}{dx}\left( xy \right)$
We know differentiation of exponential is, $\dfrac{d}{dx}\left( {{e}^{u}} \right)={{e}^{u}}.\dfrac{d}{dx}(u)$, so the above equation becomes,
$\Rightarrow {{e}^{x+y}}\dfrac{d}{dx}\left( x+y \right)=\dfrac{d}{dx}\left( xy \right)$
We know the product rule as, $\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$, applying this formula in the above equation, we get
${{e}^{x+y}}\left( 1+\dfrac{dy}{dx} \right)=x\dfrac{dy}{dx}+y\dfrac{d(x)}{dx}$
${{e}^{x+y}}\left( 1+\dfrac{dy}{dx} \right)=y+x\dfrac{dy}{dx}.$
From given expression we have${{e}^{x+y}}=xy$, putting this value in above equation, we get
$xy\left( 1+\dfrac{dy}{dx} \right)=y+x\dfrac{dy}{dx}$
$\Rightarrow xy+xy\dfrac{dy}{dx}=y+x\dfrac{dy}{dx}$
Bringing the like terms on one side, we get
$\Rightarrow xy\dfrac{dy}{dx}-x\dfrac{dy}{dx}=y-xy$
Taking out the common terms, we get
$\Rightarrow x\left( y-1 \right)\dfrac{dy}{dx}=y\left( 1-x \right)$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{y\left( 1-x \right)}{x\left( y-1 \right)}.........(i)$
Now we need to find the second order derivative, so we will differentiate the above equation with respect to $'x'$, we get
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{y\left( 1-x \right)}{x\left( y-1 \right)} \right)$
Now we know the quotient rule, i.e., $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}$, applying this formula in the above equation, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( x(y-1)\dfrac{dy}{dx}\left[ y\left( 1-x \right) \right] \right)-\left( y\left( 1-x \right)\dfrac{d}{dx}\left[ x(y-1) \right] \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
Now applying the product rule of differentiation, i.e., $\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x(y-1)\left( y\dfrac{d}{dx}\left( 1-x \right)+(1-x)\dfrac{dy}{dx} \right)-y\left( 1-x \right)\left( x\dfrac{d}{dx}\left( y-1 \right)+(y-1)\dfrac{d(x)}{dx} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
We know differentiation of constant term is zero, so solving the above equation, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x(y-1)\left( y(-1)+(1-x)\dfrac{dy}{dx} \right)-y\left( 1-x \right)\left( x\dfrac{dy}{dx}+(y-1)(1) \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
Substituting the value $\dfrac{dy}{dx}=\dfrac{y\left( 1-x \right)}{x\left( y-1 \right)}$ from equation (i) in the above equation, we get

$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x\left( y-1 \right)\left( \left( \dfrac{y\left( 1-x \right)}{x\left( y-1 \right)} \right)\left( 1-x \right)-y \right)-y\left( 1-x \right)\left( \left( y-1 \right)+x\left( \dfrac{y\left( 1-x \right)}{x\left( y-1 \right)} \right) \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
Solving the innermost brackets first, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x\left( y-1 \right)\left( \dfrac{y{{\left( 1-x \right)}^{2}}-yx(y-1)}{x\left( y-1 \right)} \right)-y\left( 1-x \right)\left( \dfrac{{{(y-1)}^{2}}+y\left( 1-x \right)}{\left( y-1 \right)} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
Cancelling the like terms, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( y-1 \right)\left( \dfrac{y{{\left( 1-x \right)}^{2}}-yx(y-1)}{\left( y-1 \right)} \right)-y\left( 1-x \right)\left( \dfrac{{{(y-1)}^{2}}+y\left( 1-x \right)}{\left( y-1 \right)} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( \dfrac{y\left( y-1 \right){{\left( 1-x \right)}^{2}}-yx{{(y-1)}^{2}}-y(1-x){{(y-1)}^{2}}-{{y}^{2}}{{\left( 1-x \right)}^{2}}}{\left( y-1 \right)} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y\left( y-1 \right){{\left( 1-x \right)}^{2}}-yx{{(y-1)}^{2}}-y(1-x){{(y-1)}^{2}}-{{y}^{2}}{{\left( 1-x \right)}^{2}}}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
Now taking $'y'$ common, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y\left[ \left( y-1 \right){{\left( 1-x \right)}^{2}}-x{{(y-1)}^{2}}-(1-x){{(y-1)}^{2}}-y{{\left( 1-x \right)}^{2}} \right]}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
Opening the two-two brackets, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y\left\{ y{{\left( 1-x \right)}^{2}}-{{\left( 1-x \right)}^{2}}-x{{(y-1)}^{2}}-{{\left( y-1 \right)}^{2}}+x{{\left( y-1 \right)}^{2}}-y{{\left( 1-x \right)}^{2}} \right\}}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
Cancelling the like terms, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y\left\{ -{{\left( 1-x \right)}^{2}}-{{\left( y-1 \right)}^{2}} \right\}}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-y\left\{ {{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}} \right\}}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
Hence proved

Note: Another way to solve this is first take log on both sides of the given expression, as shown below.
$\ln \left( {{e}^{x+y}} \right)=\ln \left( xy \right)$
$\Rightarrow xy\ln \left( e \right)=\ln \left( xy \right)$
$\Rightarrow xy=\ln \left( xy \right)$
Then perform the next steps.