# 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 :

==> 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)**

f(x) = cosx/(1+sinx). To find the differential coefficient of f(x).

We use the quotient rule for differentiating this ,as the f(x) is in the form u(x)/v(x).

Quotient rule: (u(x)/v(x))' = {u'(x)v(x)- u(x)v'(x)}/{v(x)}^2.

Here u(x) = cosx.

u' (x) = - sinx.

v(x) = 1+sinx.

v'(x) = (1+sinx)' = cosx.

Therefore , f'(x) = {cosx/(1+sinx)}'.

f'(x) = {(cosx)'(1+sinx) - (cosx)(1+sinx)'}/(1+sinx)^2.

f'(x) = {(-sinx)(1+sinx) - cosx(cosx)}/(1+sinx)^2.

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

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

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

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

Therefore -1/(1+sinx) is the diffrential coefficient of sinx/(1+sinx).