Limits and Derivatives — MATHS STD 11 Science — Question

1. $\cos9$ 

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 $\cos9$