Sum
Find `("d"y)/("d"x)`, if xy = log(xy)
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Solution
xy = log(xy)
Differentiating both sides w.r.t. x, we get
`x*("d"y)/("d"x) + y*"d"/("d"x)(x) = 1/(xy)*"d"/("d"x)(xy)`
∴ `x*("d"y)/("d"x) + y*1 = 1/(xy)[x*("d"y)/("d"x) + y*"d"/("d"x)(x)]`
∴ `x*("d"y)/("d"x) + y = 1/(xy)(x("d"y)/("d"x) + y*1)`
∴ `x*("d"y)/("d"x) + y = 1/y*("d"y)/("d"x) + 1/x`
∴ `(x - 1/y)("d"y)/("d"x) = 1/x - y`
∴ `-((1 - xy)/y)("d"y)/("d"x) = ((1 - xy)/x)`
∴ `("d"y)/("d"x) = -((1 - xy)/x) xx (y/(1 - xy))`
∴ `("d"y)/("d"x) = (-y)/x`
Concept: The Concept of Derivative - Derivatives of Logarithmic Functions
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