- #1
sponsoredwalk
- 533
- 5
Differentiate with respect to x; (using the quotient rule)
3/2x-1 (3 over 2x minus 1)
dy/dx = (2x-1)(0) - (3)(2) / (2x-1)^2
dy/dx = -6/(2x-1)^2
but my book gives -2/(2x-1)^2
now,
y = u/v and i take
u = 3 and
v = 2x-1.
dy/dx = v(du/dx) - u(dv/dx) / (v)^2
hmm...
3/2x-1 (3 over 2x minus 1)
dy/dx = (2x-1)(0) - (3)(2) / (2x-1)^2
dy/dx = -6/(2x-1)^2
but my book gives -2/(2x-1)^2
now,
y = u/v and i take
u = 3 and
v = 2x-1.
dy/dx = v(du/dx) - u(dv/dx) / (v)^2
hmm...