If f(x) = cosx/(1+sinx) find f'(x).

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f(x) = cosx / (1+ sinx)

To differentiate we will will assume that:

u= cosx    ==>  we know that  du = -sinx

v= 1+ sinx   ==>    v' = cosc

Then, f(x) = u/v

We know that , f'(x) = (u'v- uv')/v^2

Let us subsitute with u and v :

==>...

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f(x) = cosx / (1+ sinx)

To differentiate we will will assume that:

u= cosx    ==>  we know that  du = -sinx

v= 1+ sinx   ==>    v' = cosc

Then, f(x) = u/v

We know that , f'(x) = (u'v- uv')/v^2

Let us subsitute with u and v :

==> f'(x) = [-sinx (1+sinx) - (cosx*cosx)]/(1+sinx)^2

Now let us expand the brackets:

==> f'(x) = (-sinx - sin^2 x - cos^2 x)/(1+sinx)^2

              = -sinx - (sin^2 x + cos^2 x)/(1+sinx)^2

But we know that : sin^2 x + cos^2 x = 1

==> f'(x) = (-sinx -1)/(1+sinx)^2

==> f'(x) = -(sinx+1)/(1+sinx)^2

==> f'(x) = -1/(1+sinx)

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