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

Answer

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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.

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.

Last updated date: 24th Sep 2023

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