implicit diffrentiation

[kjbohn]

use implicit differrentiation to compute d^2y/dx^2 and dx/dy at (1,2) on the curve x^2+y^4=17.

i have so far:
d/dx x^2 +d/dx y^4=0
2x+2ydy/dx=0
dy/dx=-y/x=-2/1=-2
then, d^2y/dx^2=d/dx*dy/dx=d/dx(-x/y)=(-xy^-1)=-(y^1+x)(y) then i have y''=-1/y+x/y^2*y'=-1/y+(-x/y)=-1/y=x^2/y^3 and i get -5/8. The anwser in the back of my book gives me the anwser for y''=-1/16,

 

[tkhunny]

2x+2ydy/dx=0
dy/dx=-y/x=-2/1=-2

Why did you do that? Just go again.

2 + 2(product rule) = 0

You're almost done.

 

[mmm4444bot]

use implicit differrentiation to compute d^2y/dx^2 and dx/dy at (1,2) on the curve x^2+y^4=17.

dx/dy is a typo, yes?

 

[Deleted member 4993]

use implicit differrentiation to compute d^2y/dx^2 and dx/dy at (1,2) on the curve x^2+y^4=17.

i have so far:
d/dx x^2 +d/dx y^4=0

2x+2ydy/dx=0 Incorrect - it should be 4*y[sup:3ni9xdff]3[/sup:3ni9xdff]*(dy/dx)

dy/dx = -x/(2y[sup:3ni9xdff]3[/sup:3ni9xdff]) ? dx/dy = -(2y[sup:3ni9xdff]3[/sup:3ni9xdff])/x

2 + 4y[sup:3ni9xdff]3[/sup:3ni9xdff](d[sup:3ni9xdff]2[/sup:3ni9xdff]y/dx[sup:3ni9xdff]2[/sup:3ni9xdff]) + 6*y[sup:3ni9xdff]2[/sup:3ni9xdff]*(dy/dx)[sup:3ni9xdff]2[/sup:3ni9xdff] = 0

d[sup:3ni9xdff]2[/sup:3ni9xdff]y/dx[sup:3ni9xdff]2[/sup:3ni9xdff] = -[1 + 3*y[sup:3ni9xdff]2[/sup:3ni9xdff]*(dy/dx)[sup:3ni9xdff]2[/sup:3ni9xdff]]/(2y[sup:3ni9xdff]3[/sup:3ni9xdff])

and so on.....

dy/dx=-y/x=-2/1=-2

then, d^2y/dx^2=d/dx*dy/dx=d/dx(-x/y)=(-xy^-1)=-(y^1+x)(y) then i have

y''=-1/y+x/y^2*y'=-1/y+(-x/y)=-1/y=x^2/y^3

and i get -5/8. The anwser in the back of my book gives me the anwser for y''=-1/16,

 

[kjbohn]

no, its not a typo its dx/dy at (1,2)

 

[galactus]

dx/dy is the reciprocal of dy/dx.

\(\displaystyle x^{2}+y^{4}=17\)

Let's do dy/dx:

\(\displaystyle 2x+4y^{3}\frac{dy}{dx}=0\)

\(\displaystyle \frac{dy}{dx}=\frac{-x}{2y^{3}}\)

At (1,2), then we have \(\displaystyle \frac{dy}{dx}=\frac{-1}{2(2)^{3}}=\frac{-1}{16}\)

Now, let's do dx/dy:

\(\displaystyle 2x\frac{dx}{dy}+4y^{3}=0\)

\(\displaystyle \frac{dx}{dy}=\frac{-2y^{3}}{x}\)

At (1,2), \(\displaystyle \frac{dx}{dy}=\frac{-2(2)^{3}}{1}=-16\)

See?. it's just dy/dx 'flipped'. Afterall, \(\displaystyle \frac{1}{\frac{dy}{dx}}=\frac{dx}{dy}\)

 

[soroban]

Hello, kjbohn!

\(\displaystyle \text{Use implicit differrentiation to compute }\frac{d^2y}{dx^2}\text{ and }\frac{dx}{dy}\text{ at }(1,2)\text{ on the curve: }\,x^2\,+\,y^4\:=\:17\)

\(\displaystyle \text{The answer in my book gives me: }\:y'\,=\,-\tfrac{1}{16}\) .*
This is correct, but I don't know why they mentioned it.

\(\displaystyle \text{We have: }\:2x + 4y^3\frac{dy}{dx} \:=\:0 \quad\Rightarrow\quad \frac{dy}{dx} \:=\:\frac{\text{-}x}{2y^3}\)

. . \(\displaystyle \text{At }(1,2):\:\frac{dy}{dx} \,=\,-\frac{1}{16}\) .* . \(\displaystyle \Rightarrow\quad \boxed{\frac{dx}{dy} \:=\:-16}\)


\(\displaystyle \text{Then: }\:\frac{d^2y}{dx^2} \:=\:\frac{2y^3(\text{-}1) - (\text{-}x)6y^2\frac{dy}{dx}}{4y^6} \:=\:\frac{\text{-}2y^3 + 6xy^2\frac{dy}{dx}}{4y^6} \:=\:\frac{2y^2(\text{-}y + 3x\frac{dy}{dx})}{4y^6} \:=\:\frac{\text{-}y + 3x\frac{dy}{dx}}{2y^4}\)

\(\displaystyle \text{At }(1,2)\!:\;\frac{d^2y}{dx^2} \;=\;\frac{\text{-}2 + 3(1)(\text{-}\frac{1}{16})}{2(2^4)} \;=\;\frac{\text{-}2 - \frac{3}{16}}{32} \:=\:\frac{\text{-}\frac{35}{16}}{32}\)

. . \(\displaystyle \text{Therefore: }\:\boxed{\frac{d^2y}{dx^2} \;=\;-\frac{35}{512}}\)

 

[mmm4444bot]

\(\displaystyle \text{Use implicit differrentiation to compute }\frac{d^2y}{dx^2}\text{ and }\frac{dx}{dy}\text{ at }(1,2)\text{ on the curve: }\,x^2\,+\,y^4\:=\:17\)

\(\displaystyle \text{The answer in my book gives me: }\:y'\,=\,-\tfrac{1}{16}\) .*

This is correct, but I don't know why they mentioned it.

Maybe they report an answer for dy/dx because they intended to type dy/dx instead of dx/dy.

Does the book report an answer for dx/dy ?