Answer to Question #143964 in Calculus for liam donohue

Question #143964
Differentiate the following function: f(x)=ex+xe
1
Expert's answer
2020-11-16T19:57:47-0500

f(x)=ex+xexdifferentiating ex, we get exf(x) = e^x + xe^x\\ \textsf{differentiating }e^x\textsf{, we get }e^x


differentiating xex, we use the product rulelet u=xlet v=exdu=1dv=exfrom, Udv+Vduwe get xex+ex as the differentiated solution\textsf{differentiating } xe^x\textsf{, we use the product rule}\\ \textsf{let u} = x\\ \textsf{let v} = e^x\\ du = 1\\ dv = e^x\\ \textsf{from, }Udv+ Vdu\\ \textsf{we get }xe^x + e^x \textsf{ as the differentiated solution}

Substituting the differentiated value back into the original question, we have;


f(x)=(ex)+(xex+ex)f(x)=2ex+xexf(x)=ex(x+2)f'(x) = (e^x) + (xe^x + e^x)\\ f'(x) = 2e^x + xe^x\\ \therefore f'(x) = e^x(x+2)


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