Question

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

• A. $cos9$
• B. $sin9$
• C. 0
• D. 1

Answer 1

The correct option is A

$cos9$

$y=\frac{sin(x+9)}{cosx}$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{cosx\times \frac{d}{dx}sin(x+9)-sin(x+9)\times \frac{d}{dx}cosx}{co{s}^{2}x}$ (Quotient rule)

$=\frac{cosx\times cos(x+9)-sin(x+9)\times (-sinx)}{co{s}^{2}x}=\frac{cos(x+9)cosx+sin(x+9)sinx}{co{s}^{2}x}$

$=\frac{cos(x+9-x)}{co{s}^{2}x}=\frac{cos9}{co{s}^{2}x}$

Putting x = 0, we get

${\left(\frac{dy}{dx}\right)}_{x=0}=\frac{cos9}{co{s}^{2}0}=cos9(cos0=1)$

$\text{Thus},\frac{dy}{dx}atx=0iscos9$