If \[{{x}^{y}}+{{y}^{x}}={{a}^{b}},\] then find \[\frac{dy}{dx}.\] |
OR |
If \[y=\log [x+\sqrt{{{x}^{2}}+{{a}^{2}}}],\] show that \[({{x}^{2}}+{{a}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}=0.\] |
Answer:
Given equation is \[{{x}^{y}}+{{y}^{x}}={{a}^{b}}.\] Let \[u={{x}^{y}}\,\,\text{and}\,\,v={{y}^{x}}\] So, the given equation becomes \[u+v={{a}^{b}}\] On differentiating both sides w.r.t. x, we get \[\frac{du}{dx}+\frac{dv}{dx}=0\] ?(i) Now, consider \[u={{x}^{y}}\] Taking log on both sides, we get \[logu=\log \,{{x}^{y}}\] \[\Rightarrow \] \[\log u=y\log \,x\] \[[\because \,\,\,log\,{{m}^{n}}=n\,log\,m]\] On differentiating both sides w.r.t. x, we get \[\frac{1}{u}\cdot \frac{du}{dx}=y\cdot \frac{d}{dx}(\log \,x)+\log \,x\cdot \frac{d}{dx}(y)\] [by product rule of derivative] \[\Rightarrow \] \[\frac{1}{u}\cdot \frac{du}{dx}=\frac{y}{x}+\log \,x\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{du}{dx}=u\left( \frac{y}{x}+\log x\frac{dy}{dx} \right)\] \[\Rightarrow \] \[\frac{du}{dx}={{x}^{y}}\left( \frac{y}{x}+\log x\frac{dy}{dx} \right)\] \[[\because \,\,\,u={{x}^{y}}]\] \[\Rightarrow \] \[\frac{du}{dx}={{x}^{y-1}}\cdot y+{{x}^{y}}\log x\frac{dy}{dx}\] ?(ii) and \[v={{y}^{x}}\] Taking log on both sides, we get \[\log \,v=\log {{y}^{x}}\] \[\Rightarrow \] \[\log \,v=x\,\,\log y\] \[[\because \,\,\,log\,{{m}^{n}}=n\,\log \,m]\] On differentiating both sides w.r.t. x, we get \[\frac{1}{v}\cdot \frac{dv}{dx}=x\cdot \frac{d}{dx}(\log \,y)+\log \,y\cdot \frac{d}{dx}(x)\] [by product rule of derivative] \[\Rightarrow \] \[\frac{1}{v}\cdot \frac{dv}{dx}=x\cdot \frac{1}{y}\cdot \frac{dy}{dx}+\log \,y\cdot 1\] \[\Rightarrow \] \[\frac{dv}{dx}=v\left( \frac{x}{y}.\frac{dy}{dx}+\log \,y \right)\] \[\Rightarrow \] \[\frac{dv}{dx}={{y}^{x}}\left( \frac{x}{y}.\frac{dy}{dx}+\log \,y \right)\] \[[\because \,\,\,v={{y}^{x}}]\] \[\Rightarrow \] \[+\text{ }ve\] ?(iii) On putting the values from Eqs. (ii) and (iii) in Eq. (i), we get \[\left( {{x}^{y-1}}\cdot y+{{x}^{y}}\log \,x\frac{dy}{dx} \right)+\left( {{y}^{x-1}}\cdot x\frac{dy}{dx}+{{y}^{x}}\log y \right)=0\]\[\Rightarrow \] \[({{x}^{y}}\log \,x+{{y}^{x-1}}\cdot x)\frac{dy}{dx}=-\,({{y}^{x}}\log \,y+{{x}^{y-1}}\cdot y)\] \[\therefore \] \[\frac{dy}{dx}=\frac{-\,({{y}^{x}}\log y+{{x}^{y-1}}\cdot y)}{{{x}^{y}}\log x+{{y}^{x-1}}\cdot x}\] OR Consider, \[y=\log \,[x+\sqrt{{{x}^{2}}+{{a}^{2}}}]\] On differentiating both sides w.r.t. x, we get \[\frac{dy}{dx}=\frac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}\times \left[ 1+\frac{1}{2\sqrt{{{x}^{2}}+{{a}^{2}}}}\times 2x \right]\]\[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{[x+\sqrt{{{x}^{2}}+{{a}^{2}}}]}\times \left[ \frac{\sqrt{{{x}^{2}}+{{a}^{2}}}+x}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]\]\[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}={{({{x}^{2}}+{{a}^{2}})}^{-1/2}}\] ?(i) Again, differentiating both sides w.r.t.x, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-\,1}{2}{{({{x}^{2}}+{{a}^{2}})}^{-3/2}}\times 2x\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-x}{({{x}^{2}}+{{a}^{2}})\sqrt{({{x}^{2}}+{{a}^{2}})}}\] \[\Rightarrow \] \[({{x}^{2}}+{{a}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=-x\frac{dy}{dx}\] [using Eq. (i)] \[\therefore \] \[({{x}^{2}}+{{a}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}=0\] Hence proved
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