Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Mark the Correct Alternative in Each of the Following: If Y = Sin X + Cos X Sin X − Cos X Then D Y D X at X = 0 is - Mathematics

MCQ

Mark the correct alternative in each of the following:
If$y = \frac{\sin x + \cos x}{\sin x - \cos x}$ then $\frac{dy}{dx}$at x = 0 is

#### Options

• −2

•  0

• $\frac{1}{2}$

• does not exist

#### Solution

$y = \frac{\sin x + \cos x}{\sin x - \cos x}$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx} = \frac{\left( \sin x - \cos x \right) \times \frac{d}{dx}\left( \sin x + \cos x \right) - \left( \sin x + \cos x \right) \times \frac{d}{dx}\left( \sin x - \cos x \right)}{\left( \sin x - \cos x \right)^2} \left( \text{ Quotient rule } \right)$
$= \frac{\left( \sin x - \cos x \right) \times \left[ \frac{d}{dx}\left( \sin x \right) + \frac{d}{dx}\left( \cos x \right) \right] - \left( \sin x + \cos x \right) \times \left[ \frac{d}{dx}\left( \sin x \right) - \frac{d}{dx}\left( \cos x \right) \right]}{\left( \sin x - \cos x \right)^2}$
$= \frac{\left( \sin x - \cos x \right)\left( \cos x - \sin x \right) - \left( \sin x + \cos x \right)\left( \cos x + \sin x \right)}{\left( \sin x - \cos x \right)^2}$
$= \frac{- \left( \cos^2 x + \sin^2 x - 2\cos x \sin x \right) - \left( \sin^2 x + \cos^2 x + 2\sin x \cos x \right)}{\left( \sin x - \cos x \right)^2}$

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

Putting x = 0, we get

$\left( \frac{dy}{dx} \right)_{x = 0} = \frac{- 2}{\left( \sin0 - \cos0 \right)^2} = \frac{- 2}{\left( 0 - 1 \right)^2} = - 2$

Thus,

$\frac{dy}{dx}$ at x = 0 is −2.

Hence, the correct answer is option (a).

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Q 9 | Page 48