Question

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

Answer 1

\[y = \frac{\sin x + \cos x}{\sin x - \cos x}\] 

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).