Question

$\Large \sf\orange{~Question:}$
Determine the derivative of cosx/(1+sin x).
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Answer 1

Answer:

f'( x )= −1 / 1+ sin x

Step-by-step explanation:

Use the quotient rule to get the derivative. The quotient rule says that if

f ( x ) = g(x) / h(x)

the derivative f' (x) =g'(x)h(x)−g(x)h'(x) / h^2(x)

In this case g(x)=cos x and

h(x) = ( 1+sin x )

g'(x) = −sin x

h'(x)= cos x

f'(x) = −sin x (1+sinx)−cos x⋅cos x / (1+sin x)^2

f'(x) = − sin x − sin^2 x − cos^2 x / (1+sin x)2

f'(x) = − sin x + sin 2^ x + cos^2 x / (1+sin x)2

We know that sin^2 x + cos^2 x = 1 so the expression becomes

f'(x) = − sin x +1 / (1+sin x)^2

f'(x) = − 1 / 1 + sin x

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Answer 2

${\huge{\underline{\bf{\pink{Answer}}}}}$

Here  $f(x)=\frac{\cos x}{1+\sin x}$

Applying quotient rule of differentiation,

$\begin{aligned} f^{\prime}(x) &=\frac{\frac{d}{d x}(\cos x) \times(1+\sin x)-\cos x \times \frac{d}{d x}(1+\sin x)}{(1+\sin x)^{2}} \\ &=\frac{-\sin x \times(1+\sin x)-\cos x(\cos x)}{(1+\sin x)^{2}} \\ &=\frac{-\sin x-\sin ^{2} x-\cos ^{2} x}{(1+\sin x)^{2}} \\ &=\frac{-\sin x-\left(\sin ^{2} x+\cos ^{2} x\right)}{(1+\sin x)^{2}} \\ &=\frac{-\sin x-1}{(1+\sin x)^{2}}=\frac{-(1+\sin x)}{(1+\sin x)^{2}} \\ &=\frac{-1}{1+\sin x} \end{aligned}$

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