Question

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

Answer: (A)

$cos\,9$

Answer 2

We have, $y=\frac{sin\left(x+9\right)}{cos\,x}\quad\ldots\left(i\right)$ Differentiating $\left(i\right)$ w.r.t. $x$, we get $\frac{dy}{dx}=\frac{cos\,x \frac{d}{dx}\left(sin\left(x+9\right)\right)-sin\left(x+9\right) \frac{d}{dx}\left(cos\,x\right)}{cos^{2}\,x}$ $=\frac{cos\,x \left(cos\left(x+9\right)\right)\times 1 - sin\left(x+9\right)\left(- sin\, x\right)}{cos^{2}\,x}$ $=\frac{cos\, x \,cos\left(x+9\right)+sin\,x \,sin\left(x+9\right)}{cos^{2}\,x}$ $=\frac{cos\left(x-x-9\right)}{cos^{2}\,x}$ $=\frac{cos\left(-9\right)}{cos^{2}\,x}=\frac{cos\,9}{cos^{2}\,x}$ $\left(\because cos\left(-x\right)=cosx\right)$ $\therefore \frac{dy}{dx}\bigg|_{at\,x=0}$ $=\frac{cos\,9}{\left(cos\,0\right)^{2}}=\frac{cos\,9}{1}=cos\,9$