Calculating derivatives Find dy&gt;dx for the following functions. y=1−sin⁡x1+sin⁡x

Step 1: Given functiony=1−sin⁡x1+sin⁡xStep 2: To FindThe derivative of the given functionStep 3: Formula UsedDivide rule/ Quotient rule:(uv)′=u′v−v′uv2Step 4: Finding the Derivativedydx=ddx(1−sin⁡x1+sin⁡x)=(1−sin⁡x)′(1+sin⁡x)−(1+sin⁡x)′(1−sin⁡x)(1+sin⁡x)2=(−cos⁡x)(1+sin⁡x)−(cos⁡x)(1−sin⁡x)(1+sin⁡x)2=−cos⁡x−cos⁡xsin⁡x−cos⁡x+cos⁡xsin⁡x(1+sin⁡x)2=−2cos⁡x(1+sin⁡x)2

"Calculating derivatives Find dy>dx for the following functions...." solved and explained

SchachtN

Answered question

2021-10-11

Calculating derivatives Find dy>dx for the following functions.

y = \frac{1 - \sin x}{1 + \sin x}

Answer & Explanation

Leonard Stokes

Skilled2021-10-12Added 98 answers

Step 1: Given function

y = \frac{1 - \sin x}{1 + \sin x}

Step 2: To Find
The derivative of the given function
Step 3: Formula Used
Divide rule/ Quotient rule:

{(\frac{u}{v})}^{'} = \frac{u^{'} v - v^{'} u}{v^{2}}

Step 4: Finding the Derivative

\frac{d y}{d x} = \frac{d}{d x} (\frac{1 - \sin x}{1 + \sin x})

= \frac{{(1 - \sin x)}^{'} (1 + \sin x) - {(1 + \sin x)}^{'} (1 - \sin x)}{{(1 + \sin x)}^{2}}

= \frac{(- \cos x) (1 + \sin x) - (\cos x) (1 - \sin x)}{{(1 + \sin x)}^{2}}

= \frac{- \cos x - \cos x \sin x - \cos x + \cos x \sin x}{{(1 + \sin x)}^{2}}

= \frac{- 2 \cos x}{{(1 + \sin x)}^{2}}

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