Question

dx
$y=(\tan x+\sec x)(\tan x-\sec x)$
$\frac{dy}{dx}=\square$

Explanation:

Step1: Simplify the function

Use the difference - of - squares formula \((a + b)(a - b)=a^{2}-b^{2}\). Here \(a = \tan x\) and \(b=\sec x\), so \(y=\tan^{2}x-\sec^{2}x\).
Since \(\sec^{2}x = 1+\tan^{2}x\), then \(y=\tan^{2}x-(1 + \tan^{2}x)=- 1\).

Step2: Differentiate the simplified function

The derivative of a constant function \(y = C\) (where \(C=-1\)) with respect to \(x\) is given by the formula \(\frac{dy}{dx}=0\). According to the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), when \(n = 0\) (for \(y=-1=-1x^{0}\)), \(\frac{dy}{dx}=-1\times0x^{-1}=0\).

Answer:

\(0\)