At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

Differentiate. y=ln (17-x)

Sagot :

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]
[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]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.