If xmyn = (x + y)m + n, then dy/dx is:

Question

Accepted Answer

Correct answer: D — Concept: When a relation ties x and y together implicitly through products and powers, take the natural logarithm of both sides first. A logarithm turns a product into a sum and brings every exponent down as a multiplying coefficient, so the relation becomes a plain linear combination of logarithms. Differentiating that logged relation with respect to x — using d/dx(ln x) = 1/x and, by the chain rule, d/dx(ln y) = (1/y)(dy/dx) — leaves a single linear equation in dy/dx, which is solved by collecting the dy/dx terms on one side and everything else on the other. Application: Apply that method to the relation xmyn = (x + y)m + n. Take natural logarithms of both sides: m ln x + n ln y = (m + n) ln(x + y). Differentiate both sides with respect to x, treating y as a function of x: m/x + (n/y)(dy/dx) = ((m + n)/(x + y))(1 + dy/dx). Expand the right side and move every dy/dx term to the left: (n/y − (m + n)/(x + y))(dy/dx) = (m + n)/(x + y) − m/x. Combine the left-hand bracket over the common denominator y(x + y): (n(x + y) − (m + n)y)/(y(x + y)) = (nx + ny − my − ny)/(y(x + y)) = (nx − my)/(y(x + y)). Combine the right-hand side over the common denominator x(x + y): ((m + n)x − m(x + y))/(x(x + y)) = (mx + nx − mx − my)/(x(x + y)) = (nx − my)/(x(x + y)). Divide to isolate the derivative: dy/dx = ((nx − my)/(x(x + y))) × ((y(x + y))/(nx − my)). The factor (nx − my) cancels, and so does (x + y), leaving dy/dx = y/x. Cross-check: Confirm the same result a second way, without logarithms, by using the fact that both sides of the relation are homogeneous of the same degree m + n. Substitute y = cx for a constant c. The left side becomes cnxm + n and the right side becomes (1 + c)m + nxm + n, so the whole relation collapses to the condition cn = (1 + c)m + n, which fixes c and leaves x free. Every branch of the curve is therefore a straight line y = cx through the origin, and the slope of such a line is the constant c. Since c equals y/x at every point of that line, the derivative is y/x — the same value the logarithmic route produced. Contrast: Compare the remaining forms against that slope: Expression Why it is not the slope xy Homogeneous of degree 2 in x and y, while the slope of the line y = cx is the constant c, whose value does not change when x and y are scaled together; an expression of degree 2 cannot equal that slope. x/y Inverts the roles of the two variables: on the branch y = cx it equals 1/c instead of c, so it is the reciprocal of the slope rather than the slope itself. (x + y)/xy The same as 1/x + 1/y. That is what the differentiated left-hand side m/x + (n/y)(dy/dx) reduces to when m = n = 1 and dy/dx is set to 1, with the (m + n)/(x + y) term from the right-hand side left out altogether — so it comes from differentiating only one side of the relation. Result: dy/dx = y/x.

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A. (x + y)/xy

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B. xy

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C. x/y

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D. y/x

Expression	Why it is not the slope
xy	Homogeneous of degree 2 in x and y, while the slope of the line y = cx is the constant c, whose value does not change when x and y are scaled together; an expression of degree 2 cannot equal that slope.
x/y	Inverts the roles of the two variables: on the branch y = cx it equals 1/c instead of c, so it is the reciprocal of the slope rather than the slope itself.
(x + y)/xy	The same as 1/x + 1/y. That is what the differentiated left-hand side m/x + (n/y)(dy/dx) reduces to when m = n = 1 and dy/dx is set to 1, with the (m + n)/(x + y) term from the right-hand side left out altogether — so it comes from differentiating only one side of the relation.