f(x)=ex+xexdifferentiating ex, we get ex
differentiating xex, we use the product rulelet u=xlet v=exdu=1dv=exfrom, Udv+Vduwe get xex+ex as the differentiated solution
Substituting the differentiated value back into the original question, we have;
f′(x)=(ex)+(xex+ex)f′(x)=2ex+xex∴f′(x)=ex(x+2)
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