<p>Assuming that the equation determines a differentiable function f such that y = f(x), find y'.</p>
<p>5x^2 - xy - 4y^2 = 0</p>
<p>One way would be to differentiate the LHS: you would get
(5)(2x) - (x y' + y) - (4)(2y)y' = 0 using the product rule</p>
<p>Collect terms in y' to get
y'( -8y -x) = -10x + y
y' = (10 x - y) / ( 8y + x)</p>
<p>This gives you y' as a function of x and y, not just x.</p>
<p>(An alternative may be to consider the first equation as a quadratic
ay^2 + by + c = 0
Solve for y, though you would get two possible expressions for y=f(x).
You can then explicitly solve for y', for each of the two expressions; you would get y' as a function of x alone.)</p>
<p>I haven't had time to reply until now, but thanks so much!</p>
<p>Anybody have ideas for this one?</p>
<p>lim x-->2</p>
<p><a href="1/x">u</a> - (1/2)
x-2</p>
<p>The answer is -1/4, but I have no idea how to get that.</p>
<p>Combine the fractions on top to get ((2-X)/2X) / X-2
Which leads to (2-X)/2X * 1/(X-2)
Then factor out a -1 from 2-X and you get ((-1)(X-2))/2X * 1/(X-2)
The X-2 s cancel out leaving -1/2X
Plug in 2, -1/2(2) = -1/4</p>
<p>Ohhh, thank you. You're awesome.</p>