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(D) Given $f'(\ln x) = \begin{cases} 1 & 0  1 \end{cases}$.Let $\ln x = t$,then $x = e^t$.For $x > 1$,$t > 0$,so $f'(t) = e^t$.Int

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$f(x) = \begin{cases} x & x \le 0 \ e^x - 1 & x > 0 \end{cases}$

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(D) Given $f'(\ln x) = \begin{cases} 1 & 0  1 \end{cases}$.Let $\ln x = t$,then $x = e^t$.For $x > 1$,$t > 0$,so $f'(t) = e^t$.Integrating,$f(t) = e^t + C_1$.Since $f$ is continuous at $t=0$,we use the limit from the left: $f(0) = 0$.$f(0) = e^0 + C_1 = 0 \implies 1 + C_1 = 0 \implies C_1 = -1$.Thus,$f(x) = e^x - 1$ for $x > 0$.For $0 Integrating,$f(t) = t + C_2$.Given $f(0) = 0$,$0 + C_2 = 0 \implies C_2 = 0$.Thus,$f(x) = x$ for $x \le 0$.Combining these,$f(x) = \begin{cases} x & x \le 0 \ e^x - 1 & x > 0 \end{cases}$.

Let $f$ be a continuous function satisfying $f'(\ln x) = \begin{cases} 1 & 0 < x \le 1 \\ x & x > 1 \end{cases}$ and $f(0) = 0$. Then $f(x)$ can be defined as:

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