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

Step 1Given:y=sin⁡x1+cos⁡xThe given function can be written as,y=2sin⁡(x2)cos⁡(x2)2cos2(x2)(as sin⁡2x=2sin⁡xcos⁡x,cos⁡2x=2cos2x−1)y=tan⁡x2Step 2Differentiating y with respect to x,dydx=ddx(tan⁡(x2))=sec2(x2)×ddx(x2)=12sec2(x2)Hence, dydx=12sec2(x2).

Key to "Calculating derivatives Find dy>dx for the following functions.y=sin⁡x1+cos⁡x"

annlanw09y

Answered question

2022-03-25

Calculating derivatives Find dy>dx for the following functions.

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

Answer & Explanation

aznluck4u72x4

Beginner2022-03-26Added 16 answers

Step 1
Given:

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

The given function can be written as,

y = \frac{2 \sin (\frac{x}{2}) \cos (\frac{x}{2})}{2 \cos^{2} (\frac{x}{2})}

(as

\sin 2 x = 2 \sin x \cos x, \cos 2 x = 2 \cos^{2} x - 1

)

y = \tan \frac{x}{2}

Step 2
Differentiating y with respect to x,

\frac{d y}{d x} = \frac{d}{d x} (\tan (\frac{x}{2}))

= \sec^{2} (\frac{x}{2}) \times \frac{d}{d x} (\frac{x}{2})

= \frac{1}{2} \sec^{2} (\frac{x}{2})

Hence,

\frac{d y}{d x} = \frac{1}{2} \sec^{2} (\frac{x}{2})

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