What is the equation of the tangent line of #f(x) =(x+5)/(x-1)^2# at #x=-5#?

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Answer 1 · 2016-11-11T06:32:57

# L(x)= (1/36)x+5/36#

To answer this question you will need to find the derivative of:
#f(x)=(x+5)/(x-1)^2#

Use quotient rule: #f/g = (g f' - f g')/g^2#

#((x-1)^2 (1) - (x+5) 2(x-1) (1))/ ((x-1)^2)^2#

#((x-1)^2 - 2(x-1)(x+5))/ ((x-1)^4#

#(1)/(x-1)^2 - (2(x+5))/ ((x-1)^3#

#(x-1)/(x-1)^3 - (2x+10)/ ((x-1)^3#

#(x-1 - 2x-10)/ -(x-1)^3#

#f'(x)=(x+11)/ -(x-1)^3#

#f(5)=(-5+5)/(-5-1)^2=0#

#f'(x)=(-5+11)/ -(-5-1)^3= 6/216 = 1/36#

#L(x) = f(x) +f'(x)(x-a)#

#= 0 +(1/36)(x+5)#

# = (1/36)x+5/36#