Question

How do you differentiate `(cos x) / (1-sinx)`?

Answer 1

Quotient Rule:-

If `u` and `v` are two differentiable functions at `x` with `v!=0`, then `y=u/v` is differentiable at `x` and

`dy/dx=(v*du-u*dv)/v^2`

Let `y=(cosx)/(1-sinx)`

Differentiate w.r.t. 'x' using quotient rule

`implies dy/dx=((1-sinx)d/dx(cosx)-cosxd/dx(1-sinx))/(1-sinx)^2`

Since `d/dx(cosx)=-sinx` and `d/dx(1-sinx)=-cosx`

Therefore `dy/dx=((1-sinx)(-sinx)-cosx(-cosx))/(1-sinx)^2`

`implies dy/dx=(-sinx+sin^2x+cos^2x)/(1-sinx)^2`

Since `Sin^2x+Cos^2x=1`

Therefore `dy/dx=(1-sinx)/(1-sinx)^2=1/(1-Sinx)`

Hence, derivative of the given expression is `1/(1-sinx).`