Implicit Differentiation

**Truthbetold** · November 29th 2007, 04:38 PM

Find dy/dx

$y^2=\frac{x-1}{x=1}$

I got $\frac{4y}{x^2+x+1}$

The answer is $\frac{1}{y(x+1)^2}$

I believe I messed up with the fraction over a number, as in (what is the LaTex for this? It's not in the tutorial) 2/x^2+x+1/2y. I don't know how to deal with this. For some reason, perhaps a habit, I want to bring 2y to the top and multiply it by 2. After seeing the answer, I think it's the top times $\frac{1}{2y}$ .

What is it? and if I'm right about the mistake? How come you do that?

**jabroni1212** · November 29th 2007, 04:46 PM

I just understood your question. Yes when you divide the 2y it goes on the bottom not the top. Think back to dividing simple fractions. A fraction divided by a fraction you take the reciprocal of the second and multiply. ie

$\frac{\frac{1}{2}}{3}=(\frac{1}{2})(\frac{1}{3})$

I assume the denominator is x+1

Using the quotient rule on the right side

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

Simplify numerator: $2y\frac{dy}{dx}=\frac{2}{(x+1)^2}$
Solve for dy/dx: $\frac{dy}{dx}=\frac{2}{2y(x+1)^2}$
Simplify: $\frac{dy}{dx}=\frac{1}{y(x+1)^2}$

**Krizalid** · November 29th 2007, 05:11 PM

Originally Posted by Truthbetold

Find dy/dx

$y^2=\frac{x-1}{x+1}$

$\frac{{x - 1}}<br /> {{x + 1}} = \frac{{x + 1 - 2}}<br /> {{x + 1}} = 1 - \frac{2}<br /> {{x + 1}}.$

It takes less time to differentiate. (You don't even need product rule or quotient rule.)

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