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

1 Answer
Jim S
May 23, 2018

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)y=(e4ln4−e44ln2(4)−1)x−4+e4ln4−4(e4ln4−e44ln2(4)−1)

Explanation:

f(x)=e^x/lnx-xf(x)=exlnx−x, D_f=(0,1)uu(1,+oo)Df=(0,1)∪(1,+∞)

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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