<p>I am having a lot of trouble with implicit differentiation so can you guys help me find the derivatives to these 2 problems </p>
<p>(x^2)(y^2) = (x^2) + (y^2)</p>
<p>ln(xy) = x + y</p>
<p>In general, most derivatives are taken with respect to the independent variable x. However, in some cases, you can't rewrite the equation, where y is a function of x. In these cases, we can still take the derivative, but we need to indicate when we took the derivative of something other than the independent variable. The easiest way to do this is by multiplying by y' for each place where you took the derivative of a y variable.</p>
<p>Question #1: Remember that you need the product rule for the left-hand side of this equation.</p>
<p>So (x^2)(2y<em>y') + (y^2)(2x) = 2x + 2y</em>y'
Simplifying yields 2(x^2)y<em>y' + 2xy^2 = 2x + 2y</em>y'
To solve this, gather all the y' terms on one side of the equation. I tend to gather tem on the left-hand side.</p>
<p>So 2(x^2)y<em>y' - 2y</em>y' = 2x - 2xy^2
Factor out the y': y'(2(x^2)y - 2y) = 2x - 2xy^2.
And divide, so:</p>
<p>y' = (2x - 2xy^2)/(2(x^2)y - 2y).</p>
<p>Reduce by cancelling out a factor of 2 to yield.</p>
<p>y' = (x - xy^2)/((x^2)y - y).</p>
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<p>Question #2 is similar, and I think you can tackle it with the pieces that you've seen in Question #1 (remember that the derivative of ln(xy) needs the chain rule, and that chain rule yields a product rule).</p>
<p>Problems like number 1 are called implicit differentiation like you stated in your post, since neither the teacher nor the AP Test have an obligation to give you equations which are in the form of y = something (when differentiating an equation such as this, where y is expressed in terms of x, y' is already solved for the moment you differentiate). You would take the derivative of each term on both sides and solve for dy/dx, which is what TheMathProf did.</p>
<p>Also, just a bit of advice, I suggest you use dy/dx as opposed to y' when doing implicit differentiation, since amongst all the calculations the y' can begin to look like y^1, which is not good. XD</p>
<p>People would have fewer problems with y' if they would quit writing y^1. :)</p>
<p>Whether you use y' or dy/dx should be a matter of comfort. Some people are more comfortable with something that doesn't "look like a fraction". Other people tend to do silly things with y'. My personal favorite is the mistake where people rewrite y*y' as (y^2)'. :)</p>
<p>I have found with the second derivative that some people struggle extra with the dy/dx notation, but mostly because they find themselves uncomfortable with the d^2 y / dx^2 notation.</p>
<p>If you ever take multivariable calculus, you'll learn that the derivative is really a special case of the partial derivative, and the partial derivative is a special case of the directional derivative. But a neat formula for implicit differentiation involves partial derivatives and states that dy/dx= -partial x/partial y.</p>