Anthonyn2121
  • Anthonyn2121
Find dy/ds if y= s sqrt(1-s^2)+arccos(s)
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle y=s\sqrt{1-s^2}+\cos^{-1}s }\) What problem do you have with differentiating the above?
SolomonZelman
  • SolomonZelman
Can you use the *Product Rule* for the following? \(\large\color{black}{ \displaystyle s\sqrt{1-s^2} }\)
Anthonyn2121
  • Anthonyn2121
I used the product rule and derived the arccos and got |dw:1443655706621:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

SolomonZelman
  • SolomonZelman
you forgot the chain rule for \(\sqrt{1-s^2}\)
SolomonZelman
  • SolomonZelman
|dw:1443655787406:dw|
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle y=s\sqrt{1-s^2}+\cos^{-1}s }\) so the first part becomes, \(\large\color{black}{\dfrac{-s^2}{\sqrt{1-s^2}}+\sqrt{1-s^2}}\)
Anthonyn2121
  • Anthonyn2121
Thanks! I know the right answer is -2s^2/sqrt(1-s^2). Will I get the right answer if I multiply the -2s I was missing?
SolomonZelman
  • SolomonZelman
that is not right
SolomonZelman
  • SolomonZelman
you have 2 on the bottom and that cancels two on top and bottom of the first term. (this 2 on the bottom comes from the power rule of the square root)
SolomonZelman
  • SolomonZelman
Again the first term is, \(\large\color{black}{\dfrac{-s^2}{\sqrt{1-s^2}}}\)
SolomonZelman
  • SolomonZelman
then, the inverse cosine is as follows: \(\large\color{black}{ \displaystyle y=\cos^{-1}s }\) \(\large\color{black}{ \displaystyle s=\cos y }\) \(\large\color{black}{ \displaystyle 1=y'(-\sin y) }\) \(\large\color{black}{ \displaystyle -1/\sqrt{1-\cos^2y}=y' }\) \(\large\color{black}{ \displaystyle -1/\sqrt{1-s^2}=y' }\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \color{blue}{y}=\color{green}{s\sqrt{1-s^2}}+\color{red}{\cos^{-1}s} }\) \(\large\color{black}{ \displaystyle \color{blue}{dy/ds}=\color{green}{\frac{-s^2}{\sqrt{1-s^2}}+\sqrt{1-s^2}}\color{red}{-\frac{1}{\sqrt{1-s^2}}} }\)
SolomonZelman
  • SolomonZelman
questions?
Anthonyn2121
  • Anthonyn2121
Nope! I got it. Thank you so much.
SolomonZelman
  • SolomonZelman
yw

Looking for something else?

Not the answer you are looking for? Search for more explanations.