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Homework answers / question archive / Find the equation of a tangent to the curve (( x ( x - 1 ) ) / (x2 (x - 1)))3 at the point x = -1
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 = nxn-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