# f(z)Differentiate f(z)= cos z/1+ sin z

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We have to differentiate f(z)= cos z/(1+ sin z)

f(z)= cos z/(1+ sin z)

=> (cos z)(1 + sin z)^-1

f'(z) = -cos z(1 + sin z)^-2 * cos z - sin z (1 + sin z)^-1

=> -(cos z)^2/(1 + sin z)^2 - sin z/(1 + sin z)

=> [-(cos z)^2 - sin z(1 + sin z)]/(1 + sin z)^2

=> [-(cos z)^2 - sin z - (sin z)^2]/(1 + sin z)^2

=> -[1 + sin z]/(1 + sin z)^2

=> -1/(1 + sin z)

**The required derivative is -1/(1 + sin z)**

We'll re-write the expression:

f (z)= cos z/(1 + sin z), using brackets for denominator.

We notice that the expression is a quotient and we'll differentiate it using the following rule:

(u/v)' = (u'*v - u*v')/v^2

We'll put u = cos z => u' = -sin z

We'll put v = 1 + sin z => v' = cos z

We'll substitute u,v,u',v' into the expression (u/v)':

(u/v)' = [(-sin z)(1 + sin z) - (cos z)^2]/(1+sin z)^2

We'll remove the brackets:

(u/v)' = [-sin z - (sin z)^2 - (cos z)^2]/(1+sin z)^2

(u/v)' = {-sin z - [(sin z)^2 + (cos z)^2]}/(1+sin z)^2

But (sin z)^2 + (cos z)^2 = 1 (fundamental formula of trigonometry:

(u/v)' = (-sin z - 1)/(1+sin z)^2

We'll factorize by -1 the numerator:

(u/v)' = -(1+sin z)/(1+sin z)^2

We'll simplify and we'll get;

f'(z) = (u/v)' = -1/(1+sin z)^2