I will assume your equation is
(x - y - 1)^3 = x
then 3(x-y-1)^2(1 - dy/dx) = 1
at the given point...
3(1+1-1)^2(1-dy/dx) = 1
3 - 3dy/dx = 1
dy/dx = 2/3
so now you have the slope m=2/3 and the point (1,-1)
Use the method you learned in grade 9 to find the equation of that line.
Find an equation of the tangent line to the graph of the function f defined by the following equation at the indicated point.
(x - y - 1)3 = x; (1, -1)
1 answer