第12类RD Sharma解决方案–第11章区分–练习11.6

问题1.是否$y=\sqrt{x+\sqrt{x+\sqrt{x+....to\ \infin}}}$ ， 证明$\frac{dy}{dx}=\frac{1}{2y-1}$

解决方案：

We have, $y=\sqrt{x+\sqrt{x+\sqrt{x+....to\ \infin}}}$

⇒ $y=\sqrt{x+y}$

Squaring both sides, we get,

y² = x + y

$⇒2y\frac{dy}{dx}=1+\frac{dy}{dx}\\ ⇒\frac{dy}{dx}(2y-1)=1\\ ⇒\frac{dy}{dx}=\frac{1}{2y-1}$

问题2。 $y=\sqrt{cosx+\sqrt{cosx+\sqrt{cosx+ ....\ to\ \infin}}}$ ， 证明$\frac{dy}{dx}=\frac{sinx}{1-2y}$

解决方案：

We have, $y=\sqrt{cosx+\sqrt{cosx+\sqrt{cosx+ ....\ to\ \infin}}}$

⇒ $y = \sqrt{cosx+y}$

Squaring both sides, we get,

y² = cos x + y

⇒ $2y\frac{dy}{dx}=-sinx+\frac{dy}{dx}\\ ⇒ \frac{dy}{dx}(2y-1)=-sinx\\ ⇒ \frac{dy}{dx}=\frac{-sinx}{1-2y}\\ ⇒ \frac{dy}{dx}=\frac{sinx}{1-2y}$

问题3。 $y=\sqrt{logx+\sqrt{logx+\sqrt{logx+ ....\ to\ \infin}}}$ ， 证明$(2y-1)\frac{dy}{dx}=\frac{1}{x}$

解决方案：

We have, $y=\sqrt{logx+\sqrt{logx+\sqrt{logx+ ....\ to\ \infin}}}$

⇒ $y=\sqrt{logx+y}$

Squaring both sides, we get,

y² = log x + y

$⇒ 2y\frac{dy}{dx}=\frac{1}{x}+\frac{dy}{dx}\\ ⇒ \frac{dy}{dx}(2y-1)=\frac{1}{x}$

问题4。 $y=\sqrt{tanx+\sqrt{tanx+\sqrt{tanx+ ....\ to\ \infin}}}$ ， 证明$\frac{dy}{dx}=\frac{sec^2x}{2y-1}$

解决方案：

We have, $y=\sqrt{tanx+\sqrt{tanx+\sqrt{tanx+ ....\ to\ \infin}}}$

⇒ $y =\sqrt{tanx+y}$

Squaring both sides, we get,

y² = tan x + y

$⇒2y\frac{dy}{dx}=sec^2x+\frac{dy}{dx}\\ ⇒\frac{dy}{dx}(2y-1)=sec^2x\\ ⇒\frac{dy}{dx}=\frac{sec^2x}{2y-1}$

问题5。 $y=(sinx)^{(sinx)^{(sinx)^{...\infin}}}$ ， 证明$\frac{dy}{dx}=\frac{y^2cotx}{(1-y\ logsinx)}$

解决方案：

We have, $y=(sinx)^{(sinx)^{(sinx)^{...\infin}}}$

⇒ y = (sin x)^y

Taking log on both sides,

log y = log(sin x)^y

⇒ log y = y log(sin x)

$⇒\frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\{log(sinx)\}+log\ sinx\frac{dy}{dx}\\ ⇒\frac{1}{y}\frac{dy}{dx}=y\left(\frac{1}{sinx}\right)\frac{d}{dx}(sinx)+log\ sinx\frac{dy}{dx}\\ ⇒\frac{dy}{dx}\left(\frac{1}{y}-log\ sinx\right)=\frac{y}{sinx}(cosx)\\ ⇒\frac{dy}{dx}\left(\frac{1-y\ log\ sinx}{y}\right)=y\ cotx\\ ⇒\frac{dy}{dx}=\frac{y^2cotx}{1-y\ log\ sinx}$

问题6。 $y=(tanx)^{(tanx)^{(tanx)^{...\infin}}}$ ， 证明$\frac{dy}{dx}=2\ at\ x=\frac{\pi}{4}$

解决方案：

We have, $y=(tanx)^{(tanx)^{(tanx)^{...\infin}}}$

⇒ y = (tan x)^y

Taking log on both sides,

log y = log(tan x)^y

⇒ log y = y log tan x

Differentiating with respect to x using chain rule,

$⇒ \frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\{\log\ tanx\}+\log\ tan\frac{dy}{dx}\\ ⇒ \frac{1}{y}\frac{dy}{dx}=\frac{y}{tanx}\frac{d}{dx}\{tanx\}+\log\ tan\frac{dy}{dx}\\ ⇒\frac{dy}{dx}\left(\frac{1}{y}-\log tanx\right)=\frac{y}{tanx}sec^2x\\ ⇒\frac{dy}{dx}=\frac{y}{tanx}sec^2x\times\left(\frac{y}{1-y\log tanx}\right)\\$

Now,

$\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}=\frac{y\ sec^2\left(\frac{\pi}{4}\right)}{tan\left(\frac{\pi}{4}\right)}\times\frac{y}{1-y\ \log tan\left(\frac{\pi}{4}\right)}\\ ⇒\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}=\frac{y^2(\sqrt{2})^2}{1(1-y\log\ tan1)}\\ ⇒\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}=\frac{2(1)^2}{(1-0)}\ \ \ \ \ \left[\because\ (y)_{\frac{\pi}{4}}=\left(tan\frac{\pi}{4}\right)^{\left(tan\frac{\pi}{4}\right)^{\left(tan\frac{\pi}{4}\right)^{...\infin}}}=1\right]\\ ⇒\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}=2$

问题7。 $y=e^{x^{e^x}}+x^{e^{x^e}}+e^{x^{x^e}}$ ， 证明$e^{x^{e^x}}\times x^{e^{x}}\left\{\frac{e^x}{x}+e^x\times \log x\right\}+e^{e^{e^x}}\times e^{e^{^x}}\left\{\frac{1}{x}+e^x\times \log x\right\}+e^{x^{x^e}}x^{x^{e}}\times x^{e-1}\{1+e\log x\}$

解决方案：

We have, $y=e^{x^{e^x}}+x^{e^{x^e}}+e^{x^{x^e}}$

⇒ y = u + v + w

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}\ \ ...(i)$

where $u=e^{x^{e^x}},\ v=x^{e^{x^e}}\ and\ w=e^{x^{x^e}}$

Now, $u=e^{x^{e^x}}\ \ \ \ \ ....(ii)$

Taking log on both sides,

$\log u=\log e^{x^{e^x}}\\ \Rightarrow\log u=x^{e^{x}}\log e\\ \Rightarrow\log u=x^{e^{x}}\ \ \ ...(iii)\\ \log \log u=\log x^{e^{x}}\\ \Rightarrow\log \log u = e^x\log x$

Differentiating with respect to x,

$\Rightarrow\frac{1}{\log u}\frac{d}{dx}(\log u)=e^x\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(e^x)\\ \Rightarrow\frac{1}{\log u}\frac{1}{u}\frac{du}{dx}=\frac{e^x}{x}+e^x\log x\\ \Rightarrow\frac{du}{dx}=u\log u\left[\frac{e^x}{x}+e^x\log x\right]\\ \Rightarrow\frac{du}{dx}=e^{x^{e^x}}\times x^{e^{x}}\left[\frac{e^x}{x}+e^x\log x\right]\\ Now,\ v=x^{e^{x^e}}\ \ \ \ \ ...(iv)$

Taking log on both sides,

$\log v=\log x^{e^{e^x}}\\ \Rightarrow\log v = e^{e^{x}}\log x$

Taking log on both sides

$\Rightarrow\frac{1}{\log w}\frac{d}{dx}(\log w)=x^e\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x^e)\\ \Rightarrow\frac{1}{\log w}\left(\frac{1}{w}\right)\frac{dw}{dx}=x^e\left(\frac{1}{x}\right)+\log xex^{e-1}\\ \Rightarrow\frac{dw}{dx}=w\log w[x^{e-1}+e\log xx^{e-1}]\\ \Rightarrow\frac{dw}{dx}=e^{x^{x^e}}x^{x^{e}}x^{e-1}(1+e\log x)\\$

Using equation in equation (i), we get

$\frac{dy}{dx}=e^{x^{e^x}}{x^{e^x}}\left[\frac{e^x}{x}+e^x\log x\right]+x^{e^{e^x}}\times e^{e^x}\left[\frac{1}{x}+e^x\log x\right]+e^{x^{x^e}}{x^{x^e}}x^{e-1}(1+e\log x)$

问题8。 $y=(cosx)^{(cosx)^{(cosx)^\ ...\infin}}$ ， 证明$\frac{dy}{dx}=\frac{y^2tanx}{(1-y\log cosx)}$

解决方案：

We have, $y=(cosx)^{(cosx)^{(cosx)^\ ...\infin}}$

⇒ y = (cos x)^y

Taking log on both sides,

log y = log(cos x)^y

⇒ log y = y log (cos x)

Differentiating with respect to x using chain rule,

$\frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\{\log cosx\}+\log cos x\frac{dy}{dx}\\ \Rightarrow\frac{1}{y}\frac{dy}{dx}=y\left(\frac{1}{cosx}\right)\frac{d}{dx}(cosx)+\log cosx\frac{dy}{dx}\\ \Rightarrow\frac{dy}{dx}\left(\frac{1}{y}-\log cosx\right)=\frac{y}{cosx}(-sinx)\\ \Rightarrow\frac{dy}{dx}\left(\frac{1-y\log cosx}{y}\right)=-y\ tanx\\ \Rightarrow\frac{dy}{dx}=-\frac{y^2tanx}{(1-y\log cosx)}$