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

1. SolomonZelman

$$\large\color{black}{ \displaystyle y=s\sqrt{1-s^2}+\cos^{-1}s }$$ What problem do you have with differentiating the above?

2. SolomonZelman

Can you use the *Product Rule* for the following? $$\large\color{black}{ \displaystyle s\sqrt{1-s^2} }$$

3. Anthonyn2121

I used the product rule and derived the arccos and got |dw:1443655706621:dw|

4. SolomonZelman

you forgot the chain rule for $$\sqrt{1-s^2}$$

5. SolomonZelman

|dw:1443655787406:dw|

6. SolomonZelman

$$\large\color{black}{ \displaystyle y=s\sqrt{1-s^2}+\cos^{-1}s }$$ so the first part becomes, $$\large\color{black}{\dfrac{-s^2}{\sqrt{1-s^2}}+\sqrt{1-s^2}}$$

7. 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?

8. SolomonZelman

that is not right

9. 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)

10. SolomonZelman

Again the first term is, $$\large\color{black}{\dfrac{-s^2}{\sqrt{1-s^2}}}$$

11. SolomonZelman

then, the inverse cosine is as follows: $$\large\color{black}{ \displaystyle y=\cos^{-1}s }$$ $$\large\color{black}{ \displaystyle s=\cos y }$$ $$\large\color{black}{ \displaystyle 1=y'(-\sin y) }$$ $$\large\color{black}{ \displaystyle -1/\sqrt{1-\cos^2y}=y' }$$ $$\large\color{black}{ \displaystyle -1/\sqrt{1-s^2}=y' }$$

12. SolomonZelman

$$\large\color{black}{ \displaystyle \color{blue}{y}=\color{green}{s\sqrt{1-s^2}}+\color{red}{\cos^{-1}s} }$$ $$\large\color{black}{ \displaystyle \color{blue}{dy/ds}=\color{green}{\frac{-s^2}{\sqrt{1-s^2}}+\sqrt{1-s^2}}\color{red}{-\frac{1}{\sqrt{1-s^2}}} }$$

13. SolomonZelman

questions?

14. Anthonyn2121

Nope! I got it. Thank you so much.

15. SolomonZelman

yw