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Question: If\[{{e}^{x+y}}=xy\], then show that \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-y\left( {{\left( x-1 \...

Ifex+y=xy{{e}^{x+y}}=xy, then show that d2ydx2=−y((x−1)2+(y−1)2)x2(y−1)3\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}}}.

Explanation

Solution

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′'x' and ′y′'y'. Then proceed with finding the second order derivative and simplify it.

The given expression is ex+y=xy{{e}^{x+y}}=xy
Differentiate the given expression with respect to ′x′'x' , we get
ddx(ex+y)=ddx(xy)\dfrac{d}{dx}\left( {{e}^{x+y}} \right)=\dfrac{d}{dx}\left( xy \right)
We know differentiation of exponential is, ddx(eu)=eu.ddx(u)\dfrac{d}{dx}\left( {{e}^{u}} \right)={{e}^{u}}.\dfrac{d}{dx}(u), so the above equation becomes,
⇒ex+yddx(x+y)=ddx(xy)\Rightarrow {{e}^{x+y}}\dfrac{d}{dx}\left( x+y \right)=\dfrac{d}{dx}\left( xy \right)
We know the product rule as, ddx(u⋅v)=uddxv+vddxu\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
ex+y(1+dydx)=xdydx+yd(x)dx{{e}^{x+y}}\left( 1+\dfrac{dy}{dx} \right)=x\dfrac{dy}{dx}+y\dfrac{d(x)}{dx}
ex+y(1+dydx)=y+xdydx.{{e}^{x+y}}\left( 1+\dfrac{dy}{dx} \right)=y+x\dfrac{dy}{dx}.
From given expression we haveex+y=xy{{e}^{x+y}}=xy, putting this value in above equation, we get
xy(1+dydx)=y+xdydxxy\left( 1+\dfrac{dy}{dx} \right)=y+x\dfrac{dy}{dx}
⇒xy+xydydx=y+xdydx\Rightarrow xy+xy\dfrac{dy}{dx}=y+x\dfrac{dy}{dx}
Bringing the like terms on one side, we get
⇒xydydx−xdydx=y−xy\Rightarrow xy\dfrac{dy}{dx}-x\dfrac{dy}{dx}=y-xy
Taking out the common terms, we get
⇒x(y−1)dydx=y(1−x)\Rightarrow x\left( y-1 \right)\dfrac{dy}{dx}=y\left( 1-x \right)
⇒dydx=y(1−x)x(y−1).........(i)\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′'x', we get
⇒ddx(dydx)=ddx(y(1−x)x(y−1))\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., ddx(uv)=vddxu−uddxvv2\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
⇒d2ydx2=(x(y−1)dydx[y(1−x)])−(y(1−x)ddx[x(y−1)])x2(y−1)2\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., ddx(u⋅v)=uddxv+vddxu\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u, we get
⇒d2ydx2=x(y−1)(yddx(1−x)+(1−x)dydx)−y(1−x)(xddx(y−1)+(y−1)d(x)dx)x2(y−1)2\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
⇒d2ydx2=x(y−1)(y(−1)+(1−x)dydx)−y(1−x)(xdydx+(y−1)(1))x2(y−1)2\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 dydx=y(1−x)x(y−1)\dfrac{dy}{dx}=\dfrac{y\left( 1-x \right)}{x\left( y-1 \right)} from equation (i) in the above equation, we get

⇒d2ydx2=x(y−1)((y(1−x)x(y−1))(1−x)−y)−y(1−x)((y−1)+x(y(1−x)x(y−1)))x2(y−1)2\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
⇒d2ydx2=x(y−1)(y(1−x)2−yx(y−1)x(y−1))−y(1−x)((y−1)2+y(1−x)(y−1))x2(y−1)2\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
⇒d2ydx2=(y−1)(y(1−x)2−yx(y−1)(y−1))−y(1−x)((y−1)2+y(1−x)(y−1))x2(y−1)2\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}}}
⇒d2ydx2=(y(y−1)(1−x)2−yx(y−1)2−y(1−x)(y−1)2−y2(1−x)2(y−1))x2(y−1)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}}}
⇒d2ydx2=y(y−1)(1−x)2−yx(y−1)2−y(1−x)(y−1)2−y2(1−x)2x2(y−1)3\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′'y' common, we get
⇒d2ydx2=y[(y−1)(1−x)2−x(y−1)2−(1−x)(y−1)2−y(1−x)2]x2(y−1)3\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⁡(ex+y)=ln⁡(xy)\ln \left( {{e}^{x+y}} \right)=\ln \left( xy \right)
⇒xyln⁡(e)=ln⁡(xy)\Rightarrow xy\ln \left( e \right)=\ln \left( xy \right)
⇒xy=ln⁡(xy)\Rightarrow xy=\ln \left( xy \right)
Then perform the next steps.