Let g: R rightarrow R be a differentiable fun | vedclass

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(D) Differentiability of $f(x)$ at $x=0$:$	ext{LHD} = \lim_{\delta ightarrow 0^+} \frac{f(0-\delta) - f(0)}{-\delta} = \lim_{\delta ightarrow 0^+

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$(A, D)$

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(D) Differentiability of $f(x)$ at $x=0$:$	ext{LHD} = \lim_{\delta ightarrow 0^+} \frac{f(0-\delta) - f(0)}{-\delta} = \lim_{\delta ightarrow 0^+} \frac{\frac{-\delta}{|-\delta|} g(-\delta) - 0}{-\delta} = \lim_{\delta ightarrow 0^+} \frac{-g(-\delta)}{-\delta} = g^{\prime}(0) = 0$.$	ext{RHD} = \lim_{\delta ightarrow 0^+} \frac{f(0+\delta) - f(0)}{\delta} = \lim_{\delta ightarrow 0^+} \frac{\frac{\delta}{|\delta|} g(\delta) - 0}{\delta} = \lim_{\delta ightarrow 0^+} \frac{g(\delta)}{\delta} = g^{\prime}(0) = 0$.Since $	ext{LHD} = 	ext{RHD} = 0$,$f(x)$ is differentiable at $x=0$.Differentiability of $h(x) = e^{|x|}$ at $x=0$:$h(x)$ is not differentiable at $x=0$ because $|x|$ is not differentiable at $x=0$.Differentiability of $f(h(x))$ at $x=0$:Since $h(x) = e^{|x|} > 0$ for all $x$,$f(h(x)) = \frac{h(x)}{|h(x)|} g(h(x)) = 1 \cdot g(e^{|x|}) = g(e^{|x|})$.$	ext{LHD} = \lim_{\delta ightarrow 0^+} \frac{g(e^{|-\delta|}) - g(e^0)}{-\delta} = \lim_{\delta ightarrow 0^+} \frac{g(e^{\delta}) - g(1)}{-\delta} = -g^{\prime}(1)$.$	ext{RHD} = \lim_{\delta ightarrow 0^+} \frac{g(e^{\delta}) - g(1)}{\delta} = g^{\prime}(1)$.Since $g^{\prime}(1) 
eq 0$,$f(h(x))$ is not differentiable at $x=0$.Differentiability of $h(f(x))$ at $x=0$:$h(f(x)) = e^{|f(x)|} = e^{|\frac{x}{|x|} g(x)|} = e^{|g(x)|}$.$	ext{LHD} = \lim_{\delta ightarrow 0^+} \frac{e^{|g(-\delta)|} - e^{|g(0)|}}{-\delta} = \lim_{\delta ightarrow 0^+} \frac{e^{|g(-\delta)|} - 1}{|g(-\delta)|} \cdot \frac{|g(-\delta)|}{-\delta} = 1 \cdot 0 = 0$.$	ext{RHD} = \lim_{\delta ightarrow 0^+} \frac{e^{|g(\delta)|} - 1}{\delta} = \lim_{\delta ightarrow 0^+} \frac{e^{|g(\delta)|} - 1}{|g(\delta)|} \cdot \frac{|g(\delta)|}{\delta} = 1 \cdot 0 = 0$.Thus,$h(f(x))$ is differentiable at $x=0$.

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