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प्रश्न
Differentiate the following w.r.t.x. :
y = `(x log x)/(x + log x)`
उत्तर
y = `(x log x)/(x + log x)`
Differentiating w.r.t. x, we get
`("d"y)/("d"x) = "d"/("d"x) ((x log x)/(x + log x))`
= `((x + log x) "d"/("d"x) (x log x) - x log x "d"/("d"x) (x + log x))/(x + log x)^2`
= `((x + log x) (x "d"/("d"x) log x + log x "d"/("d"x) x) - x log x ("d"/("d"x) x + "d"/("d"x) log x))/(x + log x)^2`
= `((x + log x) [x (1/x) + log x (1)] - x log x (1 + 1/x))/(x + log x)^2`
= `((x + log x)(1 + log x) - x log x (1 + 1/x))/(x + log x)^2`
= `(x + x log x + log x + (log x)^2 - x log x - log x)/(x + log x)^2`
= `(x + (log x)^2)/(x + log x)^2`
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