Find y’ if xy = yx.
The Correct Answer and Explanation is:
We are tasked with finding y′y’y′ (the derivative of yyy with respect to xxx) for the equation xy=yxxy = yxxy=yx. To do so, we will first analyze the equation, differentiate implicitly, and solve for y′y’y′.
Step 1: Simplifying the equation
The given equation is:xy=yxxy = yxxy=yx
At first glance, this equation seems trivial since both sides are equal. However, to understand its implications and proceed with differentiation, we need to treat it as an implicit function.
By dividing both sides of the equation by xxx (assuming x≠0x \neq 0x=0):y=yy = yy=y
This is true by definition, so this equation does not provide us with new information. It holds for all values of xxx and yyy, meaning that any yyy satisfies this equation. However, this step does not lead us to a specific solution, so we must proceed by differentiating the original equation implicitly to find the derivative.
Step 2: Differentiating implicitly
We will now differentiate both sides of the original equation xy=yxxy = yxxy=yx with respect to xxx, treating yyy as a function of xxx.
For the left side, we apply the product rule:ddx(xy)=ddx(x)⋅y+x⋅ddx(y)=y+x⋅y′\frac{d}{dx}(xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y) = y + x \cdot y’dxd(xy)=dxd(x)⋅y+x⋅dxd(y)=y+x⋅y′
For the right side, we apply the product rule again:ddx(yx)=ddx(y)⋅x+y⋅ddx(x)=x⋅y′+y\frac{d}{dx}(yx) = \frac{d}{dx}(y) \cdot x + y \cdot \frac{d}{dx}(x) = x \cdot y’ + ydxd(yx)=dxd(y)⋅x+y⋅dxd(x)=x⋅y′+y
Now we equate both sides:y+x⋅y′=x⋅y′+yy + x \cdot y’ = x \cdot y’ + yy+x⋅y′=x⋅y′+y
Step 3: Solving for y′y’y′
We see that the terms yyy and x⋅y′x \cdot y’x⋅y′ appear on both sides of the equation, so they cancel out:y+x⋅y′−x⋅y′=yy + x \cdot y’ – x \cdot y’ = yy+x⋅y′−x⋅y′=y
This simplifies to:y=yy = yy=y
This final result confirms that the equation holds true for all xxx and yyy, and therefore y′y’y′ is undefined because there are no specific restrictions or conditions on the function yyy.
Conclusion
The equation xy=yxxy = yxxy=yx does not provide a distinct value or relationship for y′y’y′, and thus, y′y’y′ cannot be determined from the given equation alone. It is simply a tautology, holding true for any value of xxx and yyy. Therefore, y′y’y′ does not exist as a constrained function in this case.
