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fraction numerator d over denominator d x end fraction open parentheses fraction numerator sin invisible function application x over denominator 1 plus c o s invisible function application x end fraction close parentheses equals

  1. fraction numerator 1 over denominator 1 plus s i n invisible function application x end fraction    
  2. fraction numerator 1 over denominator 1 plus t a n invisible function application x end fraction    
  3. fraction numerator 1 over denominator 1 plus c o s invisible function application x end fraction    
  4. fraction numerator 1 over denominator 1 plus c o t invisible function application x end fraction    

hintHint:

We are given a function. We have to find it's derivative. We will simplify the function first. We will use different formulas for simplification.

The correct answer is: fraction numerator 1 over denominator 1 plus c o s invisible function application x end fraction


    The given function isspace y space equals fraction numerator sin x over denominator 1 space plus space cos x end fraction
    We  have to find the value of fraction numerator d y over denominator d x end fraction
    We will use these formulas to simplify the function.
    sin 2 theta space equals 2 sin theta cos theta
cos 2 theta space equals 2 cos squared theta space minus 1
    We will simplify the function
    y space equals fraction numerator sin x over denominator 1 space plus space cos x end fraction
space space space space equals fraction numerator 2 sin begin display style x over 2 end style cos begin display style x over 2 end style over denominator 2 cos squared begin display style x over 2 end style end fraction
space space space equals fraction numerator sin begin display style x over 2 end style over denominator cos begin display style x over 2 end style end fraction
space space space y equals tan x over 2
    This is the value of function in simplest form.
    We will find derivative of the function.
    fraction numerator d y over denominator d x end fraction equals fraction numerator d over denominator d x end fraction tan x over 2
space space space space space space space equals s e c squared x over 2 space fraction numerator d over denominator d x end fraction x over 2
space space space space space space space equals 1 half s e c squared x over 2
space space space space space space space equals fraction numerator 1 over denominator 2 cos squared begin display style x over 2 end style end fraction
space space space space space space equals fraction numerator 1 over denominator 1 space plus space cos x space end fraction space space space space space space space... left curly bracket space u sin g space t h e space f o r m u l a space o f space cos 2 theta right curly bracket
    Therefore, the final value is fraction numerator 1 over denominator 1 space plus space cos x end fraction

    The alternate method to solve this will be using u by method. It is method used in differentiation when we have a condition of numerator and denominator.

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