MCQ
If $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}},$ then $\frac{\text{dy}}{\text{dx}}$ at x = 0 is:
- A$\cos9$
- B$\sin9$
- C$0$
- D$1$
Solution:
$\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$
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 $\cos9$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
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