Sum

Find `dy/dx`if y = x log x (x^{2} + 1)

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#### Solution

y = x log x (x^{2} + 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)]`

= 2x^{2} log x + (x^{2} + 1) (1 + log x)

= 2x^{2} log x + (x^{2} + 1) + (x^{2} + 1) log x

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