Answer :

D3xt3R
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]
kate200468
[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]

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