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If $\operatorname{Lt}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=e^x(x+1)$ and $f(0)=0$, then $\frac{d}{d x}\left(f(x) e^{-x}\right)+\frac{d}{d x}\left(\frac{f(x)}{x}\right)=$
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The correct answer is:
$e^x+1$
Given, $\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=e^x(x+1)$
$\Rightarrow f^{\prime}(x)=e^x(x+1) \Rightarrow f(x)=x e^x+c \Rightarrow f(0)=0$
$\begin{aligned} & \therefore \quad c=0 \Rightarrow f(x)=x e^x \\ & \frac{d}{d x}\left(f(x) e^{-x}\right)+\frac{d}{d x}\left(\frac{f(x)}{x}\right) \Rightarrow \frac{d}{d x}(x)+\frac{d}{d x}\left(e^x\right)=1+e^x\end{aligned}$
$\Rightarrow f^{\prime}(x)=e^x(x+1) \Rightarrow f(x)=x e^x+c \Rightarrow f(0)=0$
$\begin{aligned} & \therefore \quad c=0 \Rightarrow f(x)=x e^x \\ & \frac{d}{d x}\left(f(x) e^{-x}\right)+\frac{d}{d x}\left(\frac{f(x)}{x}\right) \Rightarrow \frac{d}{d x}(x)+\frac{d}{d x}\left(e^x\right)=1+e^x\end{aligned}$
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