Differentiate-e-cot-x-

Question Number 29 by user1 last updated on 25/Jan/15

Differentiate e^{\sqrt{\cot x}} .

Answered by user2 last updated on 03/Nov/14

Let y = e^{\sqrt{\cot x}}

Put \cot x = t and \sqrt{\cot x} = \sqrt{t} = u, so that

y = e^{u}

\frac{d y}{d u} = e^{u}, \frac{d u}{d t} = \frac{1}{2} t^{- 1 / 2} = \frac{1}{2 \sqrt{t}}

and \frac{d t}{d x} = - {cosec}^{2} x

so, (\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}) = - \frac{1}{2} \times \frac{{cosec}^{2} x}{2 \sqrt{t}} e^{u}

= \frac{- {cosec}^{2} x}{2 \sqrt{t}} \times e \sqrt{t} [∵ u = \sqrt{t}]

= \frac{- {cosec}^{2} x}{2 \sqrt{\cot x}} \times e^{\sqrt{\cot x}}

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