# Thread: Implicit diferention

1. ## Implicit diferention

The problem is: find y'' of y = x^3 + xy - 7

the answer i got was (6x^2 + 2xy - 6x -2y)/(x-1)^3

can anyone tell me if i am right?

2. Hello, xwanderingpoetx!

I got a different result . . .

Find $\displaystyle y''$ of: .$\displaystyle y \:= \:x^3 + xy - 7$

The answer i got was: .$\displaystyle \frac{6x^2 + 2xy - 6x -2y}{(x-1)^3}$

Differentiate implicitly: .$\displaystyle \frac{dy}{dx} \:=\:3x^2 + x\frac{dy}{dx} + y \quad\Rightarrow\quad \frac{dy}{dx} - x\frac{dy}{dx} \:=\:3x^2 + y$

Factor: .$\displaystyle \frac{dy}{dx}(1-x) \:=\:3x^2+y \quad\Rightarrow\quad\boxed{ \frac{dy}{dx} \;=\;\frac{3x^2+y}{1-x}}$ .[1]

Differentiate implicitly: .$\displaystyle \frac{d^2y}{dx^2} \;=\;\frac{(1-x)\left(6x + \frac{dy}{dx}\right) - (3x^2 + y)(-1)}{(1-x)^2}$

. . $\displaystyle \frac{d^2y}{dx^2} \;=\;\frac{6x(1-x) + (1-x)\frac{dt}{dx} + (3x^2+y)}{(1-x)^2} \;=\;\frac{6x-6x^2 + 3x^2 + y + (1-x)\frac{dy}{dx}}{(1-x)^2}$

Substitute [1]: .$\displaystyle \frac{d^2y}{dx^2} \;=\;\frac{-3x^2 + 6x + y + (1-x)\frac{3x^2+y}{1-x}}{(1-x)^2} \;=\;\frac{-3x^2 + 6x + y + 3x^2 + y}{(1-x)^2}$ .**

Therefore: .$\displaystyle \boxed{\frac{d^2y}{dx^2} \;=\;\frac{6x+2y}{(1-x)^2}}$

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**

We may cancel the $\displaystyle (1-x)$ because $\displaystyle x \neq 1.$

If $\displaystyle x = 1$, the original equation becomes: .$\displaystyle y \:=\:1^3 + 1\!\cdot\!y - 7$

. . which results in the impossibility: .$\displaystyle 0 \,=\,-6$

3. Wow, thank you.

I forgot to factor out the -1 from the first derivative and I distributed the y' instead of keeping it next to the quantity (1-x)

I understand what I did wrong now, thank you.