I am finding the second derivative implicitly. For the first derivative I get dy/dx = (- 2x – 8y)/ (2y + 8x); After that, How do I get the second derivative? For each term individually, I see that (-2 - 8dy/dx)/ (2 dy/dx + 8) But I know that is not correct. How would I approach this using the division rule?
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Implicit
- Thread starter ffuh205
- Start date
Hello, ffuh205!
You should give us the entire problem . . .
I had to run the problem backwards to find the original function.
\(\displaystyle \text{I believe is it of the form: }\;x^2 + 8xy + y^2 \:=\:C,\;\text{ where }C\text{ is some constant.}\)
Differentiate implicitly:
. . \(\displaystyle 2x + 8x\frac{dy}{dx} + 8y + 2y\frac{dy}{dx} \:=\:0\)
. . . . . . . . .\(\displaystyle 8x\frac{dy}{dx} + 2y\frac{dy}{dx} \;=\;-2x-8y\)
. . . . . . . . . .\(\displaystyle 2(4x+y)\frac{dy}{dx} \;=\;-2(x + 4y)\)
. . . . . . . . . . . . . . . . \(\displaystyle \frac{dy}{dx} \;=\;\frac{-2(x+4y)}{2(4x+y)}\)
. . . . . . . . . . . . . . . . \(\displaystyle \boxed{\frac{dy}{dx} \;=\;-\frac{x+4y}{4x+y}}\)
\(\displaystyle \frac{d^2y}{dx^2} \;=\;-\left[\frac{(4x+y)(1+4\frac{dy}{dx}) - (x+4y)(4 + \frac{dy}{dx})}{(4x+y)^2}\right]\)
. . . \(\displaystyle =\;-\left[\frac{4x+ 16x\frac{dy}{dx} + y + 4y\frac{dy}{dx} - 4x - x\frac{dy}{dx} - 16y - 4y\frac{dy}{dx}}{(4x+y)^2}\right]\)
. . . \(\displaystyle =\;-\left[\frac{-15y + 15x\frac{dy}{dx}}{(4x+y)^2}\right]\)
. . . \(\displaystyle =\;15\cdot\frac{y - x\frac{dy}{dx}}{(4x+y)^2}\)
. . . \(\displaystyle =\;15\cdot\frac{y - x\left(-\frac{x+4y}{4x+y}\right)}{(4x+y)^2}\)
. . . \(\displaystyle =\;15\cdot\frac{y(4x+y) - x(-x-4y)}{(4x+y)^3}\)
. . . \(\displaystyle =\;15\cdot\frac{4xy + y^2 + x^2 + 4xy}{(4x+y)^3}\)
. . . \(\displaystyle =\; 15\cdot\frac{\overbrace{x^2+8xy + y^2}^{\text{This is }C}}{(4x+y)^3}\)
\(\displaystyle \text{Therefore: }\;\boxed{\frac{d^2y}{dx^2} \;=\;\frac{15C}{(4x+y)^3}}\)
You should give us the entire problem . . .
I am finding the second derivative implicitly.
For the first derivative I get: .\(\displaystyle \frac{dy}{dx} \:=\:\frac{- 2x - 8y}{2y + 8x}\)
After that, how do I get the second derivative? . Differentiate again . . .
For each term individually, I see that (-2 - 8dy/dx)/ (2 dy/dx + 8)
But I know that is not correct.
Of course not . . . Obviously we have to use the Quotient Rule1
I had to run the problem backwards to find the original function.
\(\displaystyle \text{I believe is it of the form: }\;x^2 + 8xy + y^2 \:=\:C,\;\text{ where }C\text{ is some constant.}\)
Differentiate implicitly:
. . \(\displaystyle 2x + 8x\frac{dy}{dx} + 8y + 2y\frac{dy}{dx} \:=\:0\)
. . . . . . . . .\(\displaystyle 8x\frac{dy}{dx} + 2y\frac{dy}{dx} \;=\;-2x-8y\)
. . . . . . . . . .\(\displaystyle 2(4x+y)\frac{dy}{dx} \;=\;-2(x + 4y)\)
. . . . . . . . . . . . . . . . \(\displaystyle \frac{dy}{dx} \;=\;\frac{-2(x+4y)}{2(4x+y)}\)
. . . . . . . . . . . . . . . . \(\displaystyle \boxed{\frac{dy}{dx} \;=\;-\frac{x+4y}{4x+y}}\)
\(\displaystyle \frac{d^2y}{dx^2} \;=\;-\left[\frac{(4x+y)(1+4\frac{dy}{dx}) - (x+4y)(4 + \frac{dy}{dx})}{(4x+y)^2}\right]\)
. . . \(\displaystyle =\;-\left[\frac{4x+ 16x\frac{dy}{dx} + y + 4y\frac{dy}{dx} - 4x - x\frac{dy}{dx} - 16y - 4y\frac{dy}{dx}}{(4x+y)^2}\right]\)
. . . \(\displaystyle =\;-\left[\frac{-15y + 15x\frac{dy}{dx}}{(4x+y)^2}\right]\)
. . . \(\displaystyle =\;15\cdot\frac{y - x\frac{dy}{dx}}{(4x+y)^2}\)
. . . \(\displaystyle =\;15\cdot\frac{y - x\left(-\frac{x+4y}{4x+y}\right)}{(4x+y)^2}\)
. . . \(\displaystyle =\;15\cdot\frac{y(4x+y) - x(-x-4y)}{(4x+y)^3}\)
. . . \(\displaystyle =\;15\cdot\frac{4xy + y^2 + x^2 + 4xy}{(4x+y)^3}\)
. . . \(\displaystyle =\; 15\cdot\frac{\overbrace{x^2+8xy + y^2}^{\text{This is }C}}{(4x+y)^3}\)
\(\displaystyle \text{Therefore: }\;\boxed{\frac{d^2y}{dx^2} \;=\;\frac{15C}{(4x+y)^3}}\)