# Find dy/dx if y=(x-1)/sin x?

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let (x-1)=u, sinx=v

dy/dx=(u(dv/dx)-v(du/dx))/(v*v)

dv/dx=1

du/dx=cosx

dy/dx=((x-1)*1)-(sinx*cosx))/(sinx*sinx)

dy/dx=((x-1)/(sinx*sinx))-tanx

To determine dy/dx, we'll have to use the quotient rule:

(f/g)' = (f'*g - f*g')/`g^(2)`

Let f(x) = x-1 => f'(x) = 1

Let g(x) = sin x => g'(x) = cos x

dy/dx = [sin x - (x-1)*cos x]/`sin^(2)` x

**Therefore, the requested first derivative of the given function is: dy/dx = (sin x + cos x - x*cos x)/`sin^(2)` x.**