# Thread: Higher Derivatives & Implicit Differentiation

1. ## Higher Derivatives & Implicit Differentiation

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.
$\sqrt{x}+\sqrt{y} = 1$

I found y'= $-\sqrt{y} /\sqrt{x}$
(that's supposed to be a fraction)

What is y''?

b) If f $(\theta)$ = cot $(\theta)$, find f'''( $\frac{\pi}{6}$)
f'(x) = $csc^2x$

What is f"? f"'?

Thank you very much!

2. Originally Posted by whiteslash
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.
$\sqrt{x}+\sqrt{y} = 1$

I found y'= $-\sqrt{y} /\sqrt{x}$
(that's supposed to be a fraction)

What is y''?

b) If f $(\theta)$ = cot $(\theta)$, find f'''( $\frac{\pi}{6}$)
f'(x) = $csc^2x$

What is f"? f"'?

Thank you very much!
$y' = -\frac{\sqrt{y}}{\sqrt{x}}
$

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

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

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

$y'' = \frac{\sqrt{x} + \sqrt{y}}{2x\sqrt{x}}$

$y'' = \frac{1}{2x\sqrt{x}}$

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

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

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

$f''{t} = 2\csc^2{t}\cot{t}$

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

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

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

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

now evaluate for $t = \frac{5\pi}{6}$

3. Here's a slightly different way to do the first problem. Since $x^{1/2}+ y^{1/2}= 1$, $(1/2)x^{-1/2}+ (1/2)y^{-1/2}y'= 0$ (from which you can get y'= -x^{-1/2}y^{1/2}= -\frac{\sqrt{y}{\sqrt{x}})

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