How do you find the derivative of #y=e^x*lnx#?

1 Answer
Jim G.
May 30, 2017

#dy/dx=e^x(lnx+1/x)#

Explanation:

#"differentiate using the "color(blue)"product rule"#

#"Given " y=g(x)h(x)" then"#

#dy/dx=g(x)h'(x)+h(x)g'(x)larr" product rule"#

#"here " g(x)=e^xrArrg'(x)=e^x#

#h(x)=lnxrArrh'(x)=1/x#

#rArrdy/dx=e^x. 1/x+lnx.e^x#

#color(white)(rArrdy/dx)=e^x(lnx+1/x)#

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