Answered

Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our 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]
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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.