Find the equation of a tangent to the curve (( x ( x - 1 ) ) / (x2 (x - 1)))3 at the point x = -1

The given curve is y = (( x ( x - 1 ) )/ (x² (x - 1)))³

since in denominator there is x and (x - 1) so x can not be equal to 0,1;

x ≠  0,1;

and by cancelling x and (x - 1 ) in numerator and denominator

the curve will become y = 1 / x³

so for tangent at x₁ = -1;

y₁ = -1 ( at x = -1)

we have to differentiate dy/dx = -3 / x⁴ ( d(xⁿ) / dx = nx^n-1)

slope of the tangent at x = -1 will be -3;

so by using point slope form of a line ;

y - y₁ = m ( x - x₁) ( here m is slope )

so equation of tangent will be y+1 = -3(x+1)

y + 1 = -3x -3;

equation of tagent :- 3x + y +4 = 0