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प्रश्न
Find the derivatives of the following:
y = `x^(logx) + (logx)^x`
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उत्तर
y = `x^(logx) + (logx)^x`
Let u = xlog x, v = (log x)x
log u = log xlog x
log u = (log x)(log x)
log u = (log x)2
`1/u * ("d"u)/("d"x) = 2 log x xx 1/x`
`("d"u)/("d"x) = 2u (logx)/x` ........(1)
v = (log x)x
log v = log (log x)x
log v = x log (log x)
`1/"v" * ("dv")/("d"x) = x xx 1/logx xx 1/x + log(log x) xx 1`
`("dv")/("d"x) = "v"[1/logx + log(logx)]` ........(2)
y = u + v
`("d"y)/("d"x) = ("d"u)/("d"x) + ("dv")/("d"x)`
`("d"y)/("d"x) = 2u logx/x + "v"[1/logx + log (log x)]`
By equation (1) and (2)
`("d"y)/("d"x) = 2 x^(logx) * logx/x + (logx)^x [1/logx + log(logx)]`
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