What is the equation of the tangent line of #f(x) =e^x/lnx-x# at #x=4#?

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Answer 1 · 2018-05-23T22:24:00

#y=(e^4/ln4-e^4/(4ln^2(4))-1)x-4+e^4/ln4-4(e^4/ln4-e^4/(4ln^2(4))-1)#

#f(x)=e^x/lnx-x#, #D_f=(0,1)uu(1,+oo)#

#f'(x)=(e^xlnx-e^x/x)/(lnx)^2-1=#

#(e^x(xlnx-1))/(x(lnx)^2)-1=#

#e^x/lnx-e^x/(xln^2x)-1#

The equation of the tangent line at #M(4,f(4))# will be

#y-f(4)=f'(4)(x-4)# #<=>#

#y-e^4/ln4+4=(e^4/ln4-e^4/(4ln^2(4))-1)(x-4)=#

#y=(e^4/ln4-e^4/(4ln^2(4))-1)x-4+e^4/ln4-4(e^4/ln4-e^4/(4ln^2(4))-1)#

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