Answer :
We have to use the chain rule's
[tex]f(x)=ln(17-x)[/tex]
[tex]f[g(x)]=ln[g(x)][/tex]
therefore
[tex]f(u)=ln(u)[/tex]
and
[tex]u=g(x)=17-x[/tex]
them we have
[tex]f'(x)=f'(u)*g'(x)[/tex]
[tex]f'(u)=\frac{1}{u}[/tex]
[tex]g'(x)=-1[/tex]
[tex]f'(x)=f'(u)*g'(x)[/tex]
[tex]f'(x)=\frac{1}{u}*(-1)[/tex]
[tex]f'(x)=-\frac{1}{u}[/tex]
[tex]\boxed{\boxed{\therefore~f'(x)=-\frac{1}{17-x}}}[/tex]
[tex]f(x)=ln(17-x)[/tex]
[tex]f[g(x)]=ln[g(x)][/tex]
therefore
[tex]f(u)=ln(u)[/tex]
and
[tex]u=g(x)=17-x[/tex]
them we have
[tex]f'(x)=f'(u)*g'(x)[/tex]
[tex]f'(u)=\frac{1}{u}[/tex]
[tex]g'(x)=-1[/tex]
[tex]f'(x)=f'(u)*g'(x)[/tex]
[tex]f'(x)=\frac{1}{u}*(-1)[/tex]
[tex]f'(x)=-\frac{1}{u}[/tex]
[tex]\boxed{\boxed{\therefore~f'(x)=-\frac{1}{17-x}}}[/tex]
[tex]y'=(17-x)'\cdot \frac{1}{ln(17-x)} =- \frac{1}{ln(17-x)} \\\\ \ \ and\ \ \ D: \ 17-x > 0\ \ \ \Rightarrow\ \ \ x<17\ \ \ \Rightarrow\ \ \ D=(17;+\infty)[/tex]
