Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.