Question

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

• A. $cos9$
• B. $sin9$
• C. $0$
• D. $1$

Answer 1

The correct option is A $cos9$

Given that

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

Using the formula of differentiation of $\frac{u}{v}$

$d\left(\frac{u}{v}\right)=\frac{vdu-udv}{v^2}$

$\Rightarrow \frac{dy}{dx}=\frac{cosx\frac{d}{dx}\left(sin\right(x+9\left)\right)-sin(x+9)\frac{d}{dx}cosx}{co{s}^{2}x}$

$\Rightarrow \frac{dy}{dx}=\frac{cosxcos(x+9)-sin(x+9)(-sinx)}{co{s}^{2}x}$

$\Rightarrow \frac{dy}{dx}=\frac{cosxcos(x+9)+sin(x+9)sinx}{co{s}^{2}x}$

substituting $x=0$, we get

$\Rightarrow \frac{dy}{dx}=\frac{cos\left(0\right)cos(0+9)+sin(0+9)sin\left(0\right)}{co{s}^2\left(0\right)}$

$\Rightarrow \frac{dy}{dx}=\frac{1\ast cos\left(9\right)+sin\left(9\right)\ast 0}{1}$

$\Rightarrow \frac{dy}{dx}=cos9$