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