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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

f(z) = cosx/ (1+ sinz)

Let f(z) = u/ v such that:

u= cosz  ==>  u' = -sinz

v = 1+sinz ==>   v' = cosx

==> f'(z) = (u'v - uv')/ v^2

               = ( -sinz(1+sinz) - (cosz*cosz)]/ (1+ sinz)^2

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

Let us factor -1 :

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

Now we know that: sin62 z + cos^2 z = 1

                   = ( -sinz - 1 ) / (1+ sinz)^2

Now we will factor -1:

                 = -(1+sinz)/ (1+sinz)62

               = -1/(1+ sinz)

Then:

f'(z) = -1/ (1+ sinz)

changchengliang's profile pic

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

Posted on

If I interpret correctly, your question should be to:

Differentiate f(z), where f(z)= cos(z) / [1+ sin(z)]

We make use of quotient rule:

f(z) = (V.dU/dz - U.dV/dz) / V^2

Let U = cos(z) and V = [1+sin(z)]

f'(z)

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

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

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

- 1 / [1+sin(z)]

 

Hence the first derivative of f(z) = f'(z) = - 1 / [1+sin(z)]

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(103 words)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(z) = cosz/(1+sinz).

The right side is in the form of u(x)/v(x).

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

Therefore ,

{(cosz)/(1+sinz)}' = {(cosz)'(1+sinz) - cosz* (1+sinz)}/(1+sinz)^2 .

{(cosz)/(1+sinz)}' = {(-sinz)(1+sinz) - cosz (cosz)}/(1+sinz)^2.

{(cosz)/(1+sinz)}' =  {-sinz -sin^2z -cos^2z}/(a+sinz)^2.

{(cosz)/(1+sinz)}' = (-sinz -(sin^2z+cos^2z)}/(1+sinz)^2.

{(cosz)/(1+sinz)}' = (-sinz-1)/(1+sinz)^2 , as cos^2z+sin^2 = 1.

{(cosz)/(1+sinz)}' = -(1+sinz)/(1+sinz)^2.

{(cosz)/(1+sinz)}' = -1/(1+sinz).

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