Prove Dy/dx = -1/(1+x)^2: A Calculus Challenge

Hey guys! Today, we're diving into a super interesting calculus problem that involves implicit differentiation and a bit of algebraic manipulation. Get ready to flex those math muscles! We're going to prove that if x√(1+y) + y√(1+x) = 0 for -1 < x < 1, then dy/dx = -1/(1+x)^2. Buckle up, it's going to be a fun ride!

The Challenge: A Deep Dive into Implicit Differentiation

Let's break down the problem statement. We're given an equation that implicitly defines y as a function of x. This means y isn't explicitly written as y = something involving x. Instead, it's mixed in with x in the equation x√(1+y) + y√(1+x) = 0. The goal is to find the derivative of y with respect to x, which is written as dy/dx, and show that it's equal to -1/(1+x)^2. This involves using the technique of implicit differentiation, where we differentiate both sides of the equation with respect to x, carefully applying the chain rule and product rule where necessary. The condition -1 < x < 1 is important because it ensures that the square roots in the original equation are well-defined (we can't take the square root of a negative number, at least not in the realm of real numbers). This constraint helps us avoid any potential issues with undefined terms during our calculations. We need to be super meticulous with our algebra to isolate dy/dx and simplify the expression to match the desired result. It's like solving a puzzle, where each step needs to be precise to fit the overall picture. So, let's roll up our sleeves and get started!

Step 1: Algebraic Gymnastics – Simplifying the Equation

Our initial equation looks a bit tangled, doesn't it? The presence of square roots makes direct differentiation a tad messy. So, the first smart move is to simplify the equation algebraically before we even think about calculus. This is a common strategy in math – making things easier for ourselves! Let's isolate one of the terms with the square root. We can rewrite the equation x√(1+y) + y√(1+x) = 0 as x√(1+y) = -y√(1+x). This sets the stage for our next move: squaring both sides. Squaring both sides will eliminate the square roots, giving us a more manageable polynomial equation. When we square both sides, we get (x√(1+y))^2 = (-y√(1+x))^2, which simplifies to x^2(1+y) = y^2(1+x). Remember, squaring both sides can sometimes introduce extraneous solutions (solutions that don't satisfy the original equation), but we'll address that later if needed. For now, let's keep moving forward. Expanding both sides of the equation, we have x^2 + x^2y = y^2 + xy^2. Now, we have a polynomial equation that's much easier to work with. Our goal is to rearrange this equation and see if we can factor it, which will help us express y in terms of x (or at least get a better handle on their relationship). Let's bring all the terms to one side: x^2 + x^2y - y^2 - xy^2 = 0. Now, let’s regroup the terms strategically to set up a factorization: (x^2 - y^2) + (x^2y - xy^2) = 0. The first group, x^2 - y^2, is a classic difference of squares, which we can factor as (x + y) (x - y). The second group, x^2y - xy^2, has a common factor of xy, so we can factor it as xy(x - y). Now our equation looks like this: (x + y) (x - y) + xy(x - y) = 0. Notice that (x - y) is a common factor in both terms! We can factor it out: (x - y) [(x + y) + xy] = 0. This gives us two possible cases: either x - y = 0 or (x + y) + xy = 0. Let's examine each case.

Step 2: Case Analysis – Unraveling the Possibilities

From our factored equation, (x - y) [(x + y) + xy] = 0, we have two possible scenarios to explore. This is where things get interesting, as we need to consider each case separately and see where it leads us. The first case is straightforward: x - y = 0. This implies that x = y. Now, let's think about this in the context of our original equation, x√(1+y) + y√(1+x) = 0. If x = y, we can substitute x for y (or vice versa) to get x√(1+x) + x√(1+x) = 0, which simplifies to 2x√(1+x) = 0. This means either x = 0 or √(1+x) = 0. If √(1+x) = 0, then 1 + x = 0, which gives us x = -1. So, in this case, we have two potential solutions: x = 0 and x = -1. However, recall that we're given the condition -1 < x < 1. This means x = -1 is not a valid solution within the given domain. Therefore, from the first case, we only have x = 0, and since x = y, this also means y = 0. This gives us one potential point (x, y) = (0, 0) to consider. The second case is a bit more involved: (x + y) + xy = 0. Our goal here is to express y explicitly as a function of x. Let's rearrange the equation to isolate y: y + xy = -x. Now, we can factor out y on the left side: y(1 + x) = -x. Finally, we can solve for y by dividing both sides by (1 + x): y = -x / (1 + x). Ah, this is a significant result! We've found an explicit expression for y in terms of x. However, we need to be cautious about the domain. Since we divided by (1 + x), we must ensure that 1 + x ≠ 0, which means x ≠ -1. This aligns with the given condition -1 < x < 1, so we're good to go. Now that we have y = -x / (1 + x), we can proceed to differentiate it with respect to x to find dy/dx. This is where the calculus magic happens!

Step 3: Calculus Time – Differentiating to Find dy/dx

Alright, we've arrived at the heart of the problem: finding dy/dx. We've successfully expressed y as an explicit function of x in the second case: y = -x / (1 + x). Now, we need to differentiate this function with respect to x. This requires using the quotient rule, which is a fundamental rule in calculus for differentiating fractions. The quotient rule states that if we have a function y = u(x) / v(x), then the derivative dy/dx is given by: dy/dx = [v(x) u'(x) - u(x) v'(x)] / [v(x)]^2. In our case, u(x) = -x and v(x) = 1 + x. Let's find their derivatives: u'(x) = -1 (the derivative of -x is -1) and v'(x) = 1 (the derivative of 1 + x is 1). Now we can plug these into the quotient rule formula: dy/dx = [(1 + x) (-1) - (-x) (1)] / (1 + x)^2. Let's simplify the numerator: (1 + x) (-1) - (-x) (1) = -1 - x + x = -1. So, our derivative becomes: dy/dx = -1 / (1 + x)^2. Boom! That's exactly what we wanted to prove. We've shown that if x√(1+y) + y√(1+x) = 0 and -1 < x < 1, then dy/dx = -1 / (1 + x)^2. But wait, we're not quite done yet. We need to consider the first case we analyzed, where x = y. In this case, we found that the only solution within the given domain is (x, y) = (0, 0). Let's see if our derivative formula holds true at this point. If we plug x = 0 into dy/dx = -1 / (1 + x)^2, we get dy/dx = -1 / (1 + 0)^2 = -1. Now, let's think about the slope of the curve x = y at the point (0, 0). Since y = x, the derivative dy/dx is simply 1. This doesn't match the value we obtained from our formula, which is -1. This discrepancy tells us something important: the case x = y doesn't represent the same function as y = -x / (1 + x) near the point (0, 0). In other words, the solution x = y is a separate branch or a singular point that doesn't follow the same derivative rule. Therefore, we can confidently conclude that for the function defined by x√(1+y) + y√(1+x) = 0 and -1 < x < 1, the derivative dy/dx is indeed given by -1 / (1 + x)^2, excluding the isolated point (0,0) which comes from the x = y case.

Conclusion: Mission Accomplished!

Woohoo! We did it! We successfully navigated the twists and turns of this calculus problem. By carefully applying algebraic manipulation, implicit differentiation, and the quotient rule, we proved that if x√(1+y) + y√(1+x) = 0 for -1 < x < 1, then dy/dx = -1/(1+x)^2. This problem beautifully illustrates the power of calculus and how different mathematical techniques can work together to solve complex challenges. Remember, the key to mastering calculus (and math in general) is practice, patience, and a willingness to explore different approaches. So, keep those brains buzzing and keep tackling those problems! You got this!