If xa .yb = (x + y ) (a + b), then show that dydx=yx - Mathematics and Statistics

Sum

If xa .yb = (x + y)^((a + b)), then show that ("d"y)/("d"x) = y/x

Solution

xa .yb = (x + y)^((a + b))

Taking logarithm of both sides, we get

log(xa.yb) = log(x + y)a+b

∴ log xa + log yb = (a + b) log (x + y)

∴ a log x + b log y = (a + b) log (x + y)

Differentiating both sides w.r.t. x, we get

"a"*1/x + "b"*1/y*("d"y)/("d"x) = ("a" + "b")*1/(x + y)*"d"/("d"x)(x + y)

∴ "a"/x + "b"/y*("d"y)/("d"x) = ("a" + "b")/(x + y)(1 + ("d"y)/("d"x))

∴ "a"/x + "b"/y*("d"y)/("d"x) = ("a" + "b")/(x + y) + ("a" + "b")/(x + y) . ("d"y)/("d"x)

∴ "b"/y*("d"y)/("d"x) - ("a" + "b")/(x + y)*("d"y)/("d"x) = ("a" + "b")/(x + y) - "a"/x

∴ ("b"/y - ("a" + "b")/(x + y)) ("d"y)/("d"x) = ("a" + "b")/(x + y) - "a"/x

∴ [("b"x + "b"y - "a"y - "b"y)/(y(x + y))]("d"y)/("d"x) = ("a"x + "b"x - "a"x - "a"y)/(x(x + y))

∴ [("b"x - "a"y)/(y(x + y))]("d"y)/("d"x) = ("b"x - "a"y)/(x(x + y))

∴ ("d"y)/("d"x) = ("b"x - "a"y)/(x(x + y)) xx (y(x + y))/("b"x - "a"y)

∴ ("d"y)/("d"x) = y/x

Concept: The Concept of Derivative - Derivatives of Logarithmic Functions
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Chapter 1.3: Differentiation - Q.5
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