Given function is \( \displaystyle y = \frac{\cos x + \sin x}{\cos x\, – \sin x} \) and we need to prove that the derivative of this function with respect to \( x \) is \(\displaystyle \sec^2\left(\frac{\pi}{4} + x\right) \).
Proof:
We are given:
\( \displaystyle y = \frac{\cos x + \sin x}{\cos x\, – \sin x} \)
Divide numerator and denominator by \( \cos x \):
\( \displaystyle y = \frac{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}\, – \frac{\sin x}{\cos x}} \\ \displaystyle = \frac{1 + \tan x}{1\, – \tan x} \\ \displaystyle = \frac{\tan \frac{\pi}{4} + \tan x}{\tan \frac{\pi}{4}\, – \tan x} \\ \)From the tangent addition formula:
\(\displaystyle \tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1\, – \tan x} \)
Therefore, \(\displaystyle y = \tan\left(\frac{\pi}{4} + x\right) \)
Differentiate using the chain rule:
\( \displaystyle \frac{dy}{dx} = \sec^2\left(\frac{\pi}{4} + x\right) \cdot \frac{d}{dx}\left(\frac{\pi}{4} + x\right) \\ \displaystyle = \sec^2\left(\frac{\pi}{4} + x\right) \cdot 1 \\ \displaystyle = \sec^2\left(\frac{\pi}{4} + x\right) \\ \)Hence, it is proved that: \(\displaystyle \boxed{\frac{dy}{dx} = \sec^2\left(\frac{\pi}{4} + x\right)} \)
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