Anthonyn2121
  • Anthonyn2121
Find dy/ds if y= s sqrt(1-s^2)+arccos(s)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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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|

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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

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