MCQ
If $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}},$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is:
  • $\cos9$
  • B
    $\sin9$
  • C
    $0$
  • D
    $1$

Answer

Correct option: A.
$\cos9$
$\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x})\frac{\text{d}}{\text{dx}}\sin(\text{x}+9)-\sin(\text{x}+9)\frac{\text{d}}{\text{dx}}(\cos\text{x})}{\cos^2\text{x}} ($Quotient rule$)$
$=\frac{(\cos\text{x})(\cos(\text{x}+9))-(\sin(\text{x}+9))(-\sin\text{x})}{\cos^2\text{x}}$
$=\frac{(\cos\text{x})(\cos(\text{x}+9))+(\sin(\text{x}+9))(\sin\text{x})}{\cos^2\text{x}}$
$=\frac{\cos(\text{x}+9-\text{x})}{\cos^2\text{x}}$
$=\frac{\cos9}{\cos^2\text{x}}$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is $\cos9$

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