MCQ

Mark the correct alternative in of the following:

If \[y = \frac{\sin\left( x + 9 \right)}{\cos x}\] then \[\frac{dy}{dx}\] at *x* = 0 is

#### Options

cos 9

sin 9

0

1

Advertisement Remove all ads

#### Solution

\[y = \frac{\sin\left( x + 9 \right)}{\cos x}\]

Differentiating both sides with respect to *x*, we get

\[\frac{dy}{dx} = \frac{\cos x \times \frac{d}{dx}\sin\left( x + 9 \right) - \sin\left( x + 9 \right) \times \frac{d}{dx}\cos x}{\cos^2 x} \left( \text{ Quotient rule } \right)\]

\[ = \frac{\cos x \times \cos\left( x + 9 \right) - \sin\left( x + 9 \right) \times \left( - \sin x \right)}{\cos^2 x}\]

\[ = \frac{\cos\left( x + 9 \right)\cos x + \sin\left( x + 9 \right)\sin x}{\cos^2 x}\]

\[ = \frac{\cos\left( x + 9 - x \right)}{\cos^2 x}\]

\[ = \frac{\cos9}{\cos^2 x}\]

\[ = \frac{\cos x \times \cos\left( x + 9 \right) - \sin\left( x + 9 \right) \times \left( - \sin x \right)}{\cos^2 x}\]

\[ = \frac{\cos\left( x + 9 \right)\cos x + \sin\left( x + 9 \right)\sin x}{\cos^2 x}\]

\[ = \frac{\cos\left( x + 9 - x \right)}{\cos^2 x}\]

\[ = \frac{\cos9}{\cos^2 x}\]

Putting

*x*= 0, we get\[\left( \frac{dy}{dx} \right)_{x = 0} = \frac{\cos9}{\cos^2 0} = \cos9\]

Thus, \[\frac{dy}{dx}\] at *x* = 0 is cos 9.

Hence, the correct answer is option (a).

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads