Problem $4\mathrm{.}37.$ We will now prove Theorem $4\mathrm{.}36.$ Let $d(x)=f(x)-g(x).$

$($a$)$ What is $d(xh)$ in terms of $f(xh)$ and $g(xh)?$

$($b$)$ What is $d(xh)-d(x)$ in terms of $f(xh),g(xh),f(x),$ and $g(x)?$

$31$

$($c$)$ Recall that the limit definition of $d^{\text{'}}(x)$ is

$d^{\text{'}}(x)=\underset{h0}{lim}\frac{d(xh)-d(x)}{h}$

Replace the numerator of this expression with your answer from part $($b$).$

$($d$)$ Rewrite your expression from part $($c$)$ as a limit of the difference of two fractions, one with $f$ terms in the numerator, and one with $g$ terms in the numerator.

$($e$)$ Now rewrite your limit as a difference of two limits$,$ one with the fraction involving the $f$ terms, the other with the fraction involving the $g$ terms.

$($f$)$ In your new expression, you have two limits$.$ One of the limits is

$\underset{h0}{lim}\frac{f(xh)-f(x)}{h}$

which is equal to $f^{\text{'}}(x).$ What is the other limit equal to$?$

$($g$)$ Finally, write $d^{\text{'}}(x)$ in terms of $f^{\text{'}}(x)$ and $g^{\text{'}}(x).$