question_answer

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