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Derivative of Some Standard Functions

If y= xtanx2,...

Question

If $y = x tan \frac{x}{2},$ then $(1 + cos x) \frac{d y}{d x} - sin x =$

A

x y

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B

y

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C

0

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D

x

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Solution

The correct option is D $x$
$y = x tan \frac{x}{2}$
$\frac{d y}{d x} = x \frac{d}{d x} (tan \frac{x}{2}) + \frac{d (x)}{d x} \cdot tan \frac{x}{2}$
$= x \cdot \frac{1}{2} {sec}^{2} \frac{x}{2} + tan \frac{x}{2}$
$= \frac{x}{2 {cos}^{2} \frac{x}{2}} + \frac{sin \frac{x}{2}}{cos \frac{x}{2}}$
$= \frac{x + 2 sin \frac{x}{2} cos \frac{x}{2}}{2 {cos}^{2} \frac{x}{2}}$
$= \frac{x + sin x}{1 + cos x}$
$\Rightarrow (1 + cos x) \frac{d y}{d x} - sin x = x$

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Derivative of Some Standard Functions

Standard XII Mathematics

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