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theEricBest ResponseYou've already chosen the best response.1
I think that you must start by simplifying. Find the derivative of (2x+y) with respect to x, and then find the derivative o f (3x+2y2) with respect to y. Let's look at the derivative you want to take with respect to x, (2x+y). Since dx is of both terms, you must take the derivative of each term with respect to x. It's just like 2x dx + y dx, right? 2x = 2(x^1) So to do the derivative, you multiply the term by x's exponent (1), and then you finish by subtract 1 from that exponent (so it's 11). (1) (2(x^(11))) = (1) (2(x^0)) = (1) (2(1)) = (1) (2) = 2. y = y(1), and 1 is the result of anything to the zeroeth power  even x to the zeroeth power. y = y(x^0) So do the same thing. You multiply the term by x's exponent (0), and then you finish by subtract 1 from that exponent (so it's 01). (0) (y(x^(01))) You can stop when you see something multiplied by "0". The whole term is obviously 0. That's what it would come down to, anyway. So 2x dx is 2 and y dx is 0. So (2x + y)dx is (2+0) = 2. For your whole equation, you must finish 2  (3x + 2y  2)dy = 0 So for (3x + 2y  2) you'll be looking at the exponents of y in each term. Do 3x dy, then 2y dy, then 2 dy. Then calculate what's left. Do that and you'll show that, by simplifying the left side of the original equation, it obviously equals the right side of the equation.
 one year ago
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