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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+2y-2) 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 1-1). (1) (2(x^(1-1))) = (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 0-1). (0) (y(x^(0-1))) 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.