Derivative Proof: Find Dy/dx If X√(1+y)+y√(1+x)=0

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Hey guys! Today, we're diving into a fun calculus problem where we need to prove a derivative given a specific condition. The problem states: If $x\sqrt{1+y}+y\sqrt{1+x}=0$ for $-1<x<1$, prove that $\frac{dy}{dx}=\frac{-1}{(1+x)^2}$. This might look a bit intimidating at first, but don't worry, we'll break it down step-by-step.

Initial Setup and Simplification

Let's start by restating the given equation: $x\sqrt{1+y}+y\sqrt{1+x}=0$. Our mission is to find $\frac{dy}{dx}$, which means we need to differentiate this equation implicitly with respect to x. Before we jump into differentiation, it's often a good idea to simplify the equation if possible. The presence of square roots suggests that squaring might help. So, let's isolate one of the terms and then square both sides.

Isolating the first term, we get:

$x\sqrt{1+y} = -y\sqrt{1+x}$

Now, squaring both sides:

$(x\sqrt{1+y})^2 = (-y\sqrt{1+x})^2$

This simplifies to:

$x^2(1+y) = y^2(1+x)$

Expanding both sides gives us:

$x^2 + x^2y = y^2 + xy^2$

This form looks a bit cleaner, and it sets us up nicely for further simplification. We've eliminated the square roots, which will make differentiation much easier. Remember, the goal here is to manipulate the equation into a form that allows us to isolate $\frac{dy}{dx}$ after differentiation.

Rearranging and Factoring

Now that we have $x^2 + x^2y = y^2 + xy^2$, let's rearrange the terms to bring similar terms together. This will help us identify potential factoring opportunities. Subtracting $y^2$ and $xy^2$ from both sides, we get:

$x^2 - y^2 + x^2y - xy^2 = 0$

Notice that we have a difference of squares ($x^2 - y^2$) and a common factor of xy in the other two terms. Let's factor these:

$(x - y)(x + y) + xy(x - y) = 0$

Now, we can see that $(x - y)$ is a common factor. Factoring it out, we have:

$(x - y)(x + y + xy) = 0$

This equation gives us two possibilities:

1. $x - y = 0$, which implies $x = y$
2. $x + y + xy = 0$

Let's consider the first case, $x = y$. If we differentiate this with respect to x, we simply get $\frac{dy}{dx} = 1$. However, this doesn't match the result we're trying to prove, which is $\frac{dy}{dx} = \frac{-1}{(1+x)^2}$. So, this case might not be the one we're interested in, or it might only be valid under specific conditions. Let's keep it in mind but focus on the second case.

The second case, $x + y + xy = 0$, looks more promising. Let's work with this equation to find $\frac{dy}{dx}$.

Implicit Differentiation

Alright, let's tackle the equation $x + y + xy = 0$. Our next step is to differentiate implicitly with respect to x. Remember, when we differentiate a term involving y with respect to x, we need to apply the chain rule, which means we'll get a $\frac{dy}{dx}$ term.

Differentiating each term with respect to x, we have:

$\frac{d}{dx}(x) + \frac{d}{dx}(y) + \frac{d}{dx}(xy) = \frac{d}{dx}(0)$

This gives us:

$1 + \frac{dy}{dx} + (x\frac{dy}{dx} + y) = 0$

Here, we used the product rule for $\frac{d}{dx}(xy)$, which states that $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$. Now we have an equation with $\frac{dy}{dx}$ terms, and our goal is to isolate $\frac{dy}{dx}$.

Isolating $\frac{dy}{dx}$

From the previous step, we have:

$1 + \frac{dy}{dx} + x\frac{dy}{dx} + y = 0$

Let's rearrange the terms to group the $\frac{dy}{dx}$ terms together:

$\frac{dy}{dx} + x\frac{dy}{dx} = -1 - y$

Now, factor out $\frac{dy}{dx}$:

$\frac{dy}{dx}(1 + x) = -1 - y$

To isolate $\frac{dy}{dx}$, we divide both sides by $(1 + x)$:

$\frac{dy}{dx} = \frac{-1 - y}{1 + x}$

We're getting closer to our desired result, but our expression for $\frac{dy}{dx}$ still contains y. We need to eliminate y using our original equation or a simplified form of it. Remember the equation we derived earlier, $x + y + xy = 0$? We can use this to express y in terms of x.

Expressing y in Terms of x

From $x + y + xy = 0$, we want to solve for y. Rearranging the terms, we get:

$y + xy = -x$

Factor out y:

$y(1 + x) = -x$

Divide by $(1 + x)$:

$y = \frac{-x}{1 + x}$

Now we have y expressed in terms of x. This is exactly what we need to substitute into our expression for $\frac{dy}{dx}$.

Substituting y into $\frac{dy}{dx}$

We have $\frac{dy}{dx} = \frac{-1 - y}{1 + x}$ and $y = \frac{-x}{1 + x}$. Let's substitute the expression for y into the equation for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-1 - (\frac{-x}{1 + x})}{1 + x}$

To simplify this, we need to get a common denominator in the numerator. Multiply -1 by $\frac{1 + x}{1 + x}$:

$\frac{dy}{dx} = \frac{\frac{-(1 + x)}{1 + x} + \frac{x}{1 + x}}{1 + x}$

Combine the fractions in the numerator:

$\frac{dy}{dx} = \frac{\frac{-1 - x + x}{1 + x}}{1 + x}$

Simplify the numerator:

$\frac{dy}{dx} = \frac{\frac{-1}{1 + x}}{1 + x}$

Now, divide by $(1 + x)$, which is the same as multiplying by $\frac{1}{1 + x}$:

$\frac{dy}{dx} = \frac{-1}{1 + x} \cdot \frac{1}{1 + x}$

Finally, we get:

$\frac{dy}{dx} = \frac{-1}{(1 + x)^2}$

Conclusion

And there you have it! We've successfully proven that if $x\sqrt{1+y}+y\sqrt{1+x}=0$ for $-1<x<1$, then $\frac{dy}{dx}=\frac{-1}{(1+x)^2}$. We started by simplifying the given equation, used implicit differentiation, solved for $\frac{dy}{dx}$, expressed y in terms of x, and substituted to get our final result. This problem is a great example of how algebraic manipulation and calculus techniques work together to solve complex problems. Keep practicing, and you'll become a pro at these in no time!