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प्रश्न
Differentiate the following w.r.t.x. :
y = `x^(3/2) "e"^xlogx`
उत्तर
y = `x^(3/2) ( "e"^xlogx)`
Differentiate with respect to x, we get
`("d"y)/("d"x) = "d"/("d"x) (x^(3/2) ("e"^x log x))`
`("d"y)/("d"x) = x^(3/2) "d"/("d"x) ("e"^x log x) + ("e"^x log x) "d"/("d"x)(x^(3/2))`
= `x^(3/2) ["e"^x "d"/("d"x) log x + log x "d"/("d"x) "e"^x] + ("e"^x log x)(3/2 x^(3/2 - 1))`
= `x^(3/2) ["e"^x (1/x) + log x ("e"^x)] + ("e"^x log x) (3/2 x^(1/2))`
= `"e"^x x^(3/2) (1/x + log x) + ("e"^x log x)(3/2 x^(1/2))`
= `x^(1/2) "e"^x[x(1/x + log x) + (log x)(3/2)]`
= `sqrt(x)"e"^x [1 + x log x + 3/2 log x]`
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