Question

Differentiate. y=ln (17-x)

Answer 1

`y'=(17-x)'\cdot (1)/(ln(17-x)) =- (1)/(ln(17-x)) \\\\ \ \ and\ \ \ D: \ 17-x > 0\ \ \ \Rightarrow\ \ \ x<17\ \ \ \Rightarrow\ \ \ D=(17;+\infty)`

Answer 2

We have to use the chain rule's


`f(x)=ln(17-x)`


`f[g(x)]=ln[g(x)]`

therefore


`f(u)=ln(u)`

and


`u=g(x)=17-x`

them we have


`f'(x)=f'(u)*g'(x)`


`f'(u)=(1)/(u)`


`g'(x)=-1`


`f'(x)=f'(u)*g'(x)`


`f'(x)=(1)/(u)*(-1)`


`f'(x)=-(1)/(u)`


`\boxed{\boxed{\therefore~f'(x)=-(1)/(17-x)}}`