1. ## Implicit diff. problem

I'm having trouble with this problem:

Solve using implicit differentiation

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

$x\cdot\frac{1}{2}(y+1)^-^\frac{1}{2}\cdot(1)=xy+1$

$x\cdot\frac{1}{2}\cdot\frac{1}{\sqrt{y+1}}=xy+1$

$\frac{x}{2\sqrt{y+1}}=xy+1$

Am I heading in the right direction? Where do I go from here?

2. Originally Posted by ascendancy523
I'm having trouble with this problem:

Solve using implicit differentiation

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

$x\cdot\frac{1}{2}(y+1)^-^\frac{1}{2}\cdot(1)=xy+1$

$x\cdot\frac{1}{2}\cdot\frac{1}{\sqrt{y+1}}=xy+1$

$\frac{x}{2\sqrt{y+1}}=xy+1$

Am I heading in the right direction? Where do I go from here?
Well, not really the right direction.

Starting with

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

Differentiate both sides with respect to x:

$\displaystyle x\cdot\frac{1}{2}(y+1)^{-\frac{1}{2}}\cdot\frac{dy}{dx}+\sqrt{y+1}=x\cdot \frac{dy}{dx}+y$

Solve for $\displaystyle\frac{dy}{dx}$ in terms of $\displaystyle x$ and $\displaystyle y$.

3. Originally Posted by undefined
Well, not really the right direction.

Starting with

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

Differentiate both sides with respect to x:

$\displaystyle x\cdot\frac{1}{2}(y+1)^{-\frac{1}{2}}\cdot\frac{dy}{dx}+\sqrt{y+1}=x\cdot \frac{dy}{dx}+y$

Solve for $\displaystyle\frac{dy}{dx}$ in terms of $\displaystyle x$ and $\displaystyle y$.
So maybe:

$x\cdot\frac{1}{2}(y+1)^-^\frac{1}{2}\cdot\frac{dy}{dx}-x\frac{dy}{dx}=-\sqrt{y+1}+y$

$\frac{dy}{dx}[x\cdot\frac{1}{2}(y+1)^-^\frac{1}{2}-x]=-\sqrt{y+1}+y$

$\frac{dy}{dx}=\frac{-\sqrt{y+1}+y}{x\cdot\frac{1}{2}(y+1)^-^\frac{1}{2}-x}$

4. Originally Posted by ascendancy523
So maybe:

$x\cdot\frac{1}{2}(y+1)^-^\frac{1}{2}\cdot\frac{dy}{dx}-x\frac{dy}{dx}=-\sqrt{y+1}+y$

$\frac{dy}{dx}[x\cdot\frac{1}{2}(y+1)^-^\frac{1}{2}-x]=-\sqrt{y+1}+y$

$\frac{dy}{dx}=\frac{-\sqrt{y+1}+y}{x\cdot\frac{1}{2}(y+1)^-^\frac{1}{2}-x}$
Looks good.

Edit: One small note: make sure this is what the problem statement asks for. You just said "solve using implicit differentiation" and I gave this answer because that is what is typically asked for.

5. Yes, this is exactly what the problem has asked for. Thank you!!!