Question

Mark the correct alternative in each of the following:

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

(a) cos 9 (b) sin 9 (c) 0 (d) 1

Answer 1

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

Differentiating both sides with respect to x, we get

$$\begin{gathered} \frac{dy}{dx}=\frac{\cos x\times {\displaystyle \frac{d}{dx}}\sin \left(x+9\right)-\sin \left(x+9\right)\times {\displaystyle \frac{d}{dx}}\cos x}{{\cos }^{2}x}\left(\mathrm{Quotient}\mathrm{rule}\right) \\ =\frac{\cos x\times \cos \left(x+9\right)-\sin \left(x+9\right)\times \left(-\sin x\right)}{{\cos }^{2}x} \\ =\frac{\cos \left(x+9\right)\cos x+\sin \left(x+9\right)\sin x}{{\cos }^{2}x} \\ =\frac{\cos \left(x+9-x\right)}{{\cos }^{2}x} \\ =\frac{\cos 9}{{\cos }^{2}x} \end{gathered}$$

Putting x = 0, we get

${\left(\frac{dy}{dx}\right)}_{x=0}=\frac{\cos 9}{{\cos }^{2}0}=\cos 9$ (cos 0 = 1)

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

Hence, the correct answer is option (a).