Math, asked by rawatanshika45127, 9 months ago

Q1) Determine the derivative of cosx/(1+sinx).



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Answers

Answered by BrainlyPopularman
67

GIVEN :

Function => cosx / (1 + sinx)

TO FIND :

Derivative = ?

SOLUTION :

• Let the function be –

\\ \bf \implies y =  \dfrac{ \cos(x) }{1 + \sin(x) }\\

• Using identity –

\\ \bf \implies \: \dfrac{d}{dx}  \left( \dfrac{u}{v} \right) = \dfrac{v. \dfrac{du}{dx}  - u.\dfrac{dv}{dx} }{ {v}^{2} } \\

• Now Differentiate with respect to 'x' –

\\ \bf \implies \dfrac{dy}{dx} =  \dfrac{  \{1 +  \sin(x) \} \dfrac{d( \cos(x)) }{dx} -  \cos(x)\dfrac{d( 1 + \sin(x)) }{dx}}{ \{ 1 + \sin(x) \}^{2}} \\

 \\\bf \implies \dfrac{dy}{dx} =  \dfrac{  \{1 +  \sin(x) \} \{ - \sin(x)  \}- \cos(x) \{ \cos(x)  \}}{ \{ 1 + \sin(x) \}^{2}  }\\

 \\\bf \implies \dfrac{dy}{dx} =  \dfrac{ -  \sin(x) -  \sin^{2} (x)   -  \cos^{2} (x)}{ \{ 1 + \sin(x) \}^{2}}\\

\\ \bf \implies \dfrac{dy}{dx} =  \dfrac{ -  \sin(x) -  \{ \sin^{2} (x)  +  \cos^{2} (x) \}}{ \{ 1 + \sin(x) \}^{2}} \\

\\ \bf \implies \dfrac{dy}{dx} =  \dfrac{ -  \sin(x) - 1}{ \{ 1 + \sin(x) \}^{2}} \\

 \\\bf \implies \dfrac{dy}{dx} =  \dfrac{- \cancel{\{1 +  \sin(x) \}}}{ \{ 1 + \sin(x) \}^{ \cancel2}} \\

 \\\bf \implies  \large{ \boxed{ \bf \dfrac{dy}{dx} =  \dfrac{-1}{1 + \sin(x)}}} \\


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Answered by ThakurRajSingh24
28

 \rm{f'(x) =   - \frac{1}{1 + sin(x)} }

Explanation :-

Applying quotient rule,

 \rm \:  \red {f(x) =  \frac{g(x)}{h(x)} }

 \implies \rm \: \large {f'(x) =  \frac{g'(x)h(x) - g(x)h'(x)}{h {}^{2}x }  ----- (1)}

In this case we have,

  • g(x) = cos(x)
  • h(x) = 1+sin(x)
  • g'(x) = -sin(x)
  • h'(x) = cos(x)

Put all values in eqn. (1) ,

 \implies \rm \: f'(x) =  \frac{ - sin \: x(1 + sin \: x) - cos \: x \times cos \: x}{(1 + sin \: x) {}^{2} }

 \implies \rm \: f'(x) =  -  \frac{  sin \: x  +  sin {}^{2} \: x  +  cos {}^{2}  \: x  }{(1 + sin \: x) {}^{2} }

 \implies \rm \: f'(x) =  -  \frac{sin (x) + 1 }{(1 + sin \: x) {}^{2} }

 \implies \rm \: f'(x) =  -  \frac{ \cancel{sin \: x + 1}}{( 1 + sin \: x) {}^{ \cancel{2}} }

 \implies \rm \: { \boxed{  \rm{ \blue{f'(x) =  -  \frac{1}{1 + sin \: x} }}}}

________________

Formula used :-

  •  \rm \: sin {}^{2}x + cos {}^{2} x = 1

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