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Anthonyn2121
 one year ago
Find dy/ds if y= s sqrt(1s^2)+arccos(s)
Anthonyn2121
 one year ago
Find dy/ds if y= s sqrt(1s^2)+arccos(s)

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4\(\large\color{black}{ \displaystyle y=s\sqrt{1s^2}+\cos^{1}s }\) What problem do you have with differentiating the above?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4Can you use the *Product Rule* for the following? \(\large\color{black}{ \displaystyle s\sqrt{1s^2} }\)

Anthonyn2121
 one year ago
Best ResponseYou've already chosen the best response.0I used the product rule and derived the arccos and got dw:1443655706621:dw

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4you forgot the chain rule for \(\sqrt{1s^2}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4dw:1443655787406:dw

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4\(\large\color{black}{ \displaystyle y=s\sqrt{1s^2}+\cos^{1}s }\) so the first part becomes, \(\large\color{black}{\dfrac{s^2}{\sqrt{1s^2}}+\sqrt{1s^2}}\)

Anthonyn2121
 one year ago
Best ResponseYou've already chosen the best response.0Thanks! I know the right answer is 2s^2/sqrt(1s^2). Will I get the right answer if I multiply the 2s I was missing?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4that is not right

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4you 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
 one year ago
Best ResponseYou've already chosen the best response.4Again the first term is, \(\large\color{black}{\dfrac{s^2}{\sqrt{1s^2}}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4then, the inverse cosine is as follows: \(\large\color{black}{ \displaystyle y=\cos^{1}s }\) \(\large\color{black}{ \displaystyle s=\cos y }\) \(\large\color{black}{ \displaystyle 1=y'(\sin y) }\) \(\large\color{black}{ \displaystyle 1/\sqrt{1\cos^2y}=y' }\) \(\large\color{black}{ \displaystyle 1/\sqrt{1s^2}=y' }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4\(\large\color{black}{ \displaystyle \color{blue}{y}=\color{green}{s\sqrt{1s^2}}+\color{red}{\cos^{1}s} }\) \(\large\color{black}{ \displaystyle \color{blue}{dy/ds}=\color{green}{\frac{s^2}{\sqrt{1s^2}}+\sqrt{1s^2}}\color{red}{\frac{1}{\sqrt{1s^2}}} }\)

Anthonyn2121
 one year ago
Best ResponseYou've already chosen the best response.0Nope! I got it. Thank you so much.
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