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After implicit differentiation, the equation becomes:
\[
\frac{y + x \frac{dy}{dx}}{2\sqrt{xy}} - \frac{y + x \frac{dy}{dx}}{xy} = 0
\]
This simplifies to two possible solutions:
1. \( \frac{dy}{dx} = -\frac{y}{x} \)
2. \( xy = 4 \)
So, the derivative \( \frac{dy}{dx} \) is either \( -\frac{y}{x} \) or the product \( xy \) equals 4.
Solución
We wish to differentiate
√(xy) = ln(xy)
with respect to x, treating y as a function of x.
Step 1. Write the equation in a clear form:
(1) √(xy) = ln(xy).
Step 2. Differentiate both sides with respect to x.
Notice that both sides involve the function u = xy. We have:
• The left‐hand side is √(xy) = [xy]^(1/2). By the chain rule,
d/dx { [xy]^(1/2) } = (1/(2√(xy))) · d/dx(xy).
• The right‐hand side is ln(xy). Again by the chain rule,
d/dx { ln(xy) } = (1/(xy)) · d/dx(xy).
Step 3. Now compute d/dx(xy).
Since x and y are both functions of x (with y = y(x)), use the product rule:
d/dx(xy) = y + x·y′.
Step 4. Substitute these derivatives back into the original differentiations.
Thus, the derivative of the left‐hand side is:
(1/(2√(xy)))(y + x y′),
and the derivative of the right‐hand side is:
(1/(xy))(y + x y′).
So our differentiated equation becomes:
(1/(2√(xy)))(y + x y′) = (1/(xy))(y + x y′).
Step 5. Factor Out the Common Term
Notice that (y + x y′) appears on both sides. Therefore we can write:
(y + x y′) · [1/(2√(xy)) – 1/(xy)] = 0.
For a product to be zero, at least one of the factors must vanish. This gives us two cases:
Case 1. Suppose
y + x y′ = 0.
Then,
y′ = – y/x.
Case 2. Suppose instead that
1/(2√(xy)) – 1/(xy) = 0.
Solve for xy:
1/(2√(xy)) = 1/(xy).
Multiply both sides by 2√(xy)·(xy) (assuming xy > 0 so that neither denominator is zero) to clear fractions:
xy = 2√(xy).
Now, let u = √(xy) so that u^2 = xy. Then the equation becomes:
u^2 = 2u.
This simplifies to:
u(u – 2) = 0.
Since u = √(xy) must be positive (note that ln(xy) is only defined when xy > 0), we have u ≠ 0. Therefore, u – 2 = 0, or u = 2. Returning to the original variables:
√(xy) = 2 ⟹ xy = 4.
Thus, if xy = 4 then the factor 1/(2√(xy)) – 1/(xy) vanishes. In this situation, the differentiated equation is automatically satisfied for any value of y′. In other words, no additional information about y′ is obtained when xy = 4.
Summary of the Answer:
The result of implicit differentiation is:
(1/(2√(xy)) – 1/(xy)) · (y + x y′) = 0.
Thus, either:
y + x y′ = 0 ⟹ y′ = – y/x,
or
xy = 4.
Either answer (or an explanation that includes the two cases) is acceptable.
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