Explanation
Let y = y_1+y_2 \Rightarrow \cfrac{dy}{dx}=\cfrac{dy_1}{dx}+\cfrac{dy_2}{dx} Now y_1 = \left(1+\frac{1}{x}\right)^x \Rightarrow \log y_1 =x \log\left(1+\frac{1}{x}\right)
Differentiating both side w.r.t x
\Rightarrow \cfrac{1}{y_1}\cfrac{dy_1}{dx} = \log(1+\frac{1}{x})+x.\cfrac{1}{1+\frac{1}{x}}.\left(\frac{-1}{x}^2\right)
\Rightarrow \cfrac{dy_1}{dx} = y_1\left[\log(1+\frac{1}{x})-\cfrac{1}{1+x}\right] And y_2 = x^{1+\frac{1}{x}} \Rightarrow \log y_2 = (1+\frac{1}{x}) \log x
Differentiating both sides,
\Rightarrow \cfrac{1}{y_2}\cfrac{dy_2}{dx} = (1+\frac{1}{x})\cfrac{1}{x}+(\log x)(\frac{-1}{x}^2)
\Rightarrow \cfrac{dy_2}{dx} = y_2\left[\cfrac{(1+x)- \log x}{x^2}\right] \therefore \cfrac{dy}{dx} = \left(1+\cfrac{1}{x}\right)\left[\log(1+\frac{1}{x})-\cfrac{1}{1+x}\right]+x^{1+\frac{1}{x}}\left[\cfrac{(1+x)- \log x}{x^2}\right]
Let \displaystyle f(x)=\begin{Bmatrix} \int _{ 0 }^{ x }{ \{ 1+|1-t| \} dt } & x>2 \\ 5x-7 & x\le 2 \end{Bmatrix}
Let \displaystyle y=\int _{ 0 }^{ x }{ \{ 1+|1-t| \} dt } as x>2
\displaystyle y=\int _{ 0 }^{ 1 }{ (1+1-t)dt } +\int _{ 1 }^{ x }{ tdt }
\displaystyle y=2-\frac { 1 }{ 2 } +\frac { { (x }^{ 2 }-1) }{ 2 }
\displaystyle y=1+\frac{ { x }^{ 2 } }{ 2 }
\displaystyle f(x)=\begin{Bmatrix} 1+\frac { { x }^{ 2 } }{ 2 } ,\quad \quad & x>2 \\ 5x-7,\quad \quad & x\le 2 \end{Bmatrix}
\displaystyle f({ 2 }^{ + })=1+\frac { 4 }{ 2 } =3
\displaystyle f({ 2 }^{ - })=10-7=3
Differentiating with respect to x we get
\displaystyle { f }^{ 1 }(x)=\begin{Bmatrix} x; & x>2 \\ 5; & x\le 2 \end{Bmatrix}
We can see that it is not differentiable at x=2 but continuous at x=2
\displaystyle y\left ( n \right )=e^{x+x^{2}+...+x^{n}}=e^{\dfrac{x\left ( 1-x^{n} \right )}{1-x}} So \displaystyle \dfrac{dy\left ( n \right )}{dx}=e^{\dfrac{x\left ( 1-x^{n} \right )}{1-x}}\times \dfrac{d}{dx}\left ( \dfrac{x\left ( 1-x^{n} \right )}{1-x} \right ) \displaystyle \lim_{n\rightarrow \infty }\dfrac{dy\left ( n \right )}{dx}=\lim_{n}e^{\dfrac{x}{1-x}-\dfrac{x^{n+1}}{1-x}}\dfrac{d}{dx}\left [ \lim_{n\rightarrow \infty }\dfrac{x}{1-x}-\dfrac{x^{n+1}}{1-x} \right ] \displaystyle =e^{\dfrac{x}{1-x}}\dfrac{d}{dx}\left ( \dfrac{x}{1-x} \right ) \displaystyle =e^{\dfrac{x}{1-x}}\dfrac{1}{\left ( 1-x \right )^{2}} \displaystyle \lim_{n\rightarrow \infty }\dfrac{dy\left ( n \right )}{dx}|_{x=\frac{1}{2}}=4e
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