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

1 Answer
Jul 21, 2016

y=-9/5(x+1)

Explanation:

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)