Hi MHF,

There are two questions that are confusing me. I understand the material, but when I was working them out, I would get stuck (with algebra parts) and I do not know how to continue.

a) Find y'' by implicit differentiation.

\(\displaystyle \sqrt{x}+\sqrt{y} = 1\)

I found y'=\(\displaystyle -\sqrt{y} /\sqrt{x}\)

(that's supposed to be a fraction)

What is y''?

b) If f\(\displaystyle (\theta)\) = cot\(\displaystyle (\theta)\), find f'''(\(\displaystyle \frac{\pi}{6}\))

f'(x) = \(\displaystyle csc^2x\)

What is f"? f"'?

Thank you very much!

\(\displaystyle y' = -\frac{\sqrt{y}}{\sqrt{x}}

\)

\(\displaystyle y'' = -\frac{\sqrt{x} \cdot \frac{y'}{2\sqrt{y}} - \sqrt{y} \cdot \frac{1}{2\sqrt{x}}}{x}

\)

\(\displaystyle y'' = \frac{-\frac{1}{2} - \frac{\sqrt{y}}{2\sqrt{x}}}{x}

\)

\(\displaystyle y'' = -\frac{2\sqrt{x}}{2\sqrt{x}} \cdot \frac{-\frac{1}{2} - \frac{\sqrt{y}}{2\sqrt{x}}}{x}

\)

\(\displaystyle y'' = \frac{\sqrt{x} + \sqrt{y}}{2x\sqrt{x}}\)

\(\displaystyle y'' = \frac{1}{2x\sqrt{x}}\)

\(\displaystyle f(t) = \cot{t}

\)

\(\displaystyle f'(t) = -\csc^2{t}\)

\(\displaystyle f''(t) = -2\csc{t} \cdot (-\csc{t}\cot{t})\)

\(\displaystyle f''{t} = 2\csc^2{t}\cot{t}\)

\(\displaystyle f''{t} = 2(\cot^2{t} + 1)\cot{t}\)

\(\displaystyle f''(t) = 2(\cot{t})^3+2\cot{t}\)

\(\displaystyle f'''(t) = 6(\cot{t})^2(-\csc^2{t}) - 2\csc^2{t}\)

\(\displaystyle f'''(t) = -2\csc^2{t}(3\cot^2{t} + 1) \)

now evaluate for \(\displaystyle t = \frac{5\pi}{6}\)