Let $f$ and $g$ be differentiable functions with the following properties:

$$\begin{array}{c l} (\text{i}) & g(x) \gt 0 \text{ for all } x \\ (\text{ii}) & f(0) = 1 \end{array}$$

If $h(x)=f(x)g(x) \text{ and } h'(x)=f(x)g'(x)$, then $f(x) =$

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In general, for $h(x)=f(x)g(x)$, we would use the product rule to obtain:

$$h'(x)=f(x)g'(x)+f'(x)g(x)$$

Notice that the question specifies that $h'(x)$ is only equal to the first part of our derivative. For this to be true, the second part has to be equal to zero:

$$\colorbox{yellow}{$h'(x)=f(x)g'(x)$}+\colorbox{chartreuse}{$f'(x)g(x)$}$$ $$\colorbox{chartreuse}{$f'(x)g(x) = 0$} \text { if } h'(x)=f(x)g'(x)$$

The question specifies that $g(x)$ is always positive, so $f'(x)$ must equal 0 so that the entire expression is 0.

$$\colorbox{chartreuse}{$f'(x)g(x) = 0$}$$ $$f'(x)=0 \text{ if } g(x)>0$$

If $f'(x)=0$, the graph of $f$ must be a horizontal line.

Since $f(0)=1$, $f(x)=1$ for all values $x$.