I think my teacher was talking about turning y into y(x) or something but I'm just completely lost on this one

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- Mar 25th 2008, 07:12 PMlyzzi3Implicit Differentiation
I think my teacher was talking about turning y into y(x) or something but I'm just completely lost on this one

- Mar 25th 2008, 07:23 PMTheEmptySet
$\displaystyle 5x^2+4x+xy=2$ using implicit we get

$\displaystyle 10x+4+y+xy'=0$

$\displaystyle y(2)=-13$

Solving for y' we get

$\displaystyle y'=- \left( \frac{10x+4+y}{x} \right) $

evaluate at x=2

$\displaystyle y'(2)=-\left( \frac{10(2)+4+(-13)}{2} \right)=-\frac{11}{2}=5.5$ - Mar 25th 2008, 07:26 PMo_O
Let's do an example to illustrate implicit differentiation:

$\displaystyle x^{2} + y^{3} = 9$

Now, if you were to ask to find the derivative of this, you*could*solve for y then differentiate the equation with respect to x. However, not all functions are easily represented only in terms of x (ex. you could not solve for y for this equation: $\displaystyle x^{2} + xy + y^{2} = 5$). Fortunately, we are still able to find the derivative to such functions.

Note that y is a function of x so that we can differentiate it normally. However, we would have to use the chain rule. I think if you study the example, you'll get what I mean:

$\displaystyle x^{2} + y^{3} = 9$

$\displaystyle 2x + \underbrace{3y^{2}y'}_{} = 0 \quad \mbox{Differentiated with respect to x}$

Looking at the underbrace, notice how we used the power rule (3y^2) and in addition the chain rule (hence the y'). Continuing on to solve for y':

$\displaystyle 3y^{2}y' = -2x$

$\displaystyle y' = \frac{-2x}{3y^{2}}$

Note it's perfectly fine to have both x and y in our solution to y'.

That is basically the trick when implicitly differentiating equations. See if you can extend this to your question. - Mar 25th 2008, 07:33 PMSengNee