Sum
Differentiate the following w. r. t. x. : ex log x
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Solution
Let y = ex log x
Differentiating w.r.t. x, we get
`dy/dx = d/dx ("e"^xlog x)`
= `"e"^xd/dx(logx) + (logx) d/dx ("e"^x)`
=`"e"^x(1/x) + (logx)("e"^x)`
= `"e"^x(1/x+logx)`
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