find-the-derivative-of-sec-1-cos-x-

Question Number 24913 by chernoaguero@gmail.com last updated on 28/Nov/17

find the derivative of

(\sec \sqrt{1 + \cos x)}

Commented by prakash jain last updated on 28/Nov/17

u = 1 + \cos x

v = \sqrt{1 + \cos x}

\frac{d}{d x} \sec (\sqrt{1 + \cos x})

= \frac{d}{d x} \sec v

= \frac{d}{d v} \sec v \cdot \frac{d}{d x} v

= \frac{d}{d v} \sec v \cdot \frac{d}{d x} \sqrt{u}

= \frac{d}{d v} \sec v \cdot \frac{d}{d u} \sqrt{u} \cdot \frac{d}{d x} u

= \frac{d}{d v} \sec v \cdot \frac{d}{d u} \sqrt{u} \cdot \frac{d}{d x} (1 + \cos x)

= \tan v \cdot \sec v \cdot \frac{1}{2 \sqrt{u}} (- \sin x)

= \tan \sqrt{1 + \cos x} \cdot \sec \sqrt{1 + \cos x} \cdot \frac{1}{2 \sqrt{1 + \cos x}} (- \sin x)

Commented by chernoaguero@gmail.com last updated on 28/Nov/17

Thank alot sir