Explanation
$$\log _{e}(x+\log _{e}^{(x+…)})$$
$$\therefore y= \log _{e}^{x+y}$$ $$\because $$ (as goes on to infinity)
$$e^{y} = x+y$$
Differentiate on both sides
$$e^{y}\dfrac{dy}{dx}= 1+ \dfrac{dy}{dx}$$
$$\because \dfrac{dy}{dx}= \dfrac{1}{e^{y}-1}$$
$$\because at (x=e^{2}-2, y=\sqrt {2}) =\dfrac{1}{e^{\sqrt{2}}-1}$$
Let $$y = 10^{x^{x}}+x^{x^{10}} +x^{10^{x}}$$
Let $$ h=10^{x^{x}}$$
Differentiate on both sides w.r.t $$x$$
$$\cfrac{dh}{dx} =10^{x^{x}} \log (10).\cfrac{d(x^{x})}{dx}$$
$$\cfrac{dh}{dx} =10^{x^{x}} x^{x}(\log x + 1 )$$ $$ (\because log 10 =1)$$
Let $$t = x^{x^{10}}$$
Apply log on both sides
$$ \log t = x^{10} \log x$$
Differentiate on both sides with respect to $$x$$
$$ \cfrac{1}{t}\cfrac{dt}{dx}=10x^{9} \log x + x^{9}$$
$$\cfrac{dt}{dx}=x^{x^{10}}x^{9}(1+ 10\log x )$$
Let $$m= x^{10^{x}}$$
$$\log m=10^{x}\log x$$
Differentiate both sides w.r.t $$x$$
$$\cfrac{1}{m}\cfrac{dm}{dx}=(10^{x} \log 10)\log x +\cfrac{10^{x}}{x}$$
$$\therefore \cfrac{dm}{dx}= x^{10^{x}}(10^{x})\left(\cfrac{1}{x}+ \log x\right)$$
$$ y= h+t+m$$
Differentiating on both sides
$$\cfrac{dy}{dx}=\cfrac{dh}{dx}+\cfrac{dt}{dx}+\cfrac{dm}{dx}$$
$$\therefore \cfrac{dy}{dx} =10^{x^{x}}x^{x} (\log x + 1 )+ x^{10^{x}}x^{9}\left(1+10 \log x\right)+ x^{10^{x}}10^{x}\left(\cfrac{1}{x}+ \log x\right)$$
We have,
$$ y={{\left( {{x}^{x}} \right)}^{x}} $$
$$ y={{x}^{{{x}^{2}}}} $$
On taking $$\log $$ both sides, we get
$$\log y={{x}^{2}}\log x$$ …… (1)
On differentiating w.r.t $$x$$, we get
$$ \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{x}+\log x\left( 2x \right) $$
$$ \dfrac{1}{y}\dfrac{dy}{dx}=x+2x\log x $$
$$ \dfrac{dy}{dx}=y\left( x+2x\log x \right) $$
$$ \dfrac{dy}{dx}={{\left( {{x}^{x}} \right)}^{x}}\left( x+2x\log x \right) $$
$$ \dfrac{dy}{dx}=x{{\left( {{x}^{x}} \right)}^{x}}\left( 1+2\log x \right) $$
Hence, this is the answer.
Equation of given curve is
$${{\left( \dfrac{x}{a} \right)}^{n}}+{{\left( \dfrac{y}{b} \right)}^{n}}=2$$
Differentiation both side with respect to x and we get,
$$ \dfrac{n}{a}{{\left( \dfrac{x}{a} \right)}^{n-1}}+\dfrac{n}{b}{{\left( \dfrac{y}{b} \right)}^{n-1}}\dfrac{dy}{dx}=0 $$
$$ \Rightarrow \dfrac{dy}{dx}=\dfrac{-\dfrac{n}{a}{{\left( \dfrac{x}{a} \right)}^{n-\grave{\ }}}}{\dfrac{n}{b}{{\left( \dfrac{y}{b} \right)}^{n-1}}} $$
$$ \Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{\left( a,b \right)}}=\dfrac{-b}{a} $$
Then, equation of tangent at point (a, b) is,
$$ \left( y-b \right)=\dfrac{-b}{a}\left( x-a \right) $$
$$ \Rightarrow ay-ab=-bx+ab $$
$$ \Rightarrow bx+ay=2ab $$
$$ \Rightarrow \dfrac{x}{a}+\dfrac{y}{b}=2 $$
Thus, the given curve touches the line
$$\dfrac{x}{a}+\dfrac{y}{b}=2$$
At point $$\left( a,\,b \right)$$ $$\forall \,n\in N$$
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