What is the equation of the tangent line of #f(x)=x^2-lnx^2/(x^2+9x)-1# at #x=-1#?

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Answer 1 · 2016-07-21T02:59:14

#y=-9/5(x+1)#

#f(x) = x^2 - lnx^2/(x^2+9) -1#

#f'(x) = 2x - {(x^2+9)* 1/x^2*2x - lnx^2*2x}/(x^2+9)^2-0#
(Power Rule, Standard differential, Chain Rule and Quotient Rule)

Replacing #x=-1# in both of the above:

#f(-1) = 1 -Ln(1)/10 -1# #= 1 -0 -1 = 0#

#f'(-1) = -2 -{10*-2/1 -0}/10^2#
#= -2 - (-20/100) = -2+2/10 = -18/10 = -9/5#

Hence the tangent touches the curve at the point #(-1,0)# and has a slope of #-9/5#

The equation of a tangent at the point#(x_1, y_1)# with slope #m# is given by: #(y-y_1) = m(x-x_1)#

Therefore the tangent in this case has the equation:

#(y-0) = -9/5(x-(-1))#
#y=-9/5(x+1)#