Higher Derivatives & Implicit Differentiation

May 2010
1
0
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!
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
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 \(\displaystyle t = \frac{5\pi}{6}\)
 
  • Like
Reactions: whiteslash

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Here's a slightly different way to do the first problem. Since \(\displaystyle x^{1/2}+ y^{1/2}= 1\), \(\displaystyle (1/2)x^{-1/2}+ (1/2)y^{-1/2}y'= 0\) (from which you can get y'= -x^{-1/2}y^{1/2}= -\frac{\sqrt{y}{\sqrt{x}})

Differentiating \(\displaystyle x^{-1/2}+ y^{-1/2}y'= 0\) with respect to x, \(\displaystyle -(1/2)x^{-3/2}- (1/2)y^{-3/2}y'+ y^{-1/2}y"= 0\) so that \(\displaystyle y^{1/2}y"= (1/2)(x^{-3/2}- y^{-3/2}y')\), \(\displaystyle y"= (1/2)(x^{-3/2}y^{-1/2}- y^{-5/2}y'= (1/2)(x^{-3/2}y^{-1/2}+x^{-1/2}y^{-2}\)