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Question
If xm . yn = (x + y)m+n, then show that `"dy"/"dx" = y/x`
Solution
xm . yn = (x + y)m+n
Taking logarithm on both sides we get,
m log x + n log y = (m + n) log(x + y)
Differentiating with respect to x,
`m1/x + n1/y "dy"/"dx" = (m + n) 1/(x + y)(1 + "dy"/"dx")`
`"dy"/"dx"(n/y - (m + n)/(x + y)) = (m + n)/(x + y) - m/x`
`"dy"/"dx" ((nx + ny - my - ny)/(y(x + y))) = ((mx + nx - mx - my)/(x(x + y)))`
`"dy"/"dx" ((nx - my)/(y(x + y))) = (nx - my)/(x(x + y))`
`"dy"/"dx" = y/x`
Hence proved.
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