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Question
Find `("d"y)/("d"x)`, if y = xx + (7x – 1)x
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Solution
y = xx + (7x – 1)x
Let u = xx and v = (7x – 1)x
∴ y = u + v
Differentiating both sides w.r.t. x, we get
`("d"y)/("d"x) = "du"/("d"x) + "dv"/("d"x)` ......(i)
Now, u = xx
Taking logarithm of both sides, we get
log u = log (xx)
∴ log u = x. log x
Differentiating both sides w.r.t. x, we get
`"d"/("d"x)(log "u") = x*"d"/("d"x)(log x) + logx*"d"/("d"x)(x)`
∴ `1/"u"*"du"/("d"x) = x*1/x + logx*1`
∴ `1/"u"*"du"/("d"x)` = 1 + log x
∴ `"du"/("d"x)` = u(1 + log x)
∴ `"du"/("d"x)` = xx(1 + log x) ......(ii)
Also, v = (7x – 1)x
Taking logarithm of both sides, we get
log v = log(7x – 1)x
∴ log v = x.log(7x – 1)
Differentiating both sides w.r.t. x, we get
`"d"/("d"x)(log "v") = x*"d"/("d"x)[log(7x - 1)] + log(7x - 1)*"d"/("d"x)(x)`
∴ `1/"v"*"dv"/("d"x) = x*1/(7x - 1)*"d"/("d"x)(7x- 1) + log(7x - 1)*1`
∴ `1/"v"*"dv"/("d"x) = x/(7x - 1)(7 - 0) + log(7x - 1)`
∴ `"dv"/("d"x) = "v"[(7x)/(7x - 1) + log(7x - 1)]`
∴ `"dv"/("d"x) = (7x - 1)^x[(7x)/(7x - 1) + log(7x - 1)]` .....(iii)
Substituting (ii) and (iii) in (i), we get
`("d"y)/("d"x) = x^x(1 + logx) + (7x - 1)^x[log(7x - 1) + (7x)/(7x - 1)]`
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