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
Verified
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
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.
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