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Question
Find `dy/dx`if y = x log x (x2 + 1)
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Solution
y = x log x (x2 + 1)
Differentiating w.r.t. x, we get
`dy/dx = d/dx(x)(logx)(x^2 + 1)`
= `(x)(logx)d/dx(x^2 + 1) - (x^2 + 1)d/dx((x)(logx))`
= `(xlogx)(2x + 0) + (x^2 + 1)[xd/dx(logx) + (logx)d/dx(x)]`
=`2x^2logx + (x^2 + 1)[x xx 1/x + (logx)(1)]`
= 2x2 log x + (x2 + 1) (1 + log x)
= 2x2 log x + (x2 + 1) + (x2 + 1) log x
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