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Derive this with respect to x
$$\frac{3}{(1 + secx) }$$

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$$~~~~\dfrac{d}{dx} \left(\dfrac{3}{1 + \sec x} \right)\\\\\\=3 \dfrac{d}{dx} \left( \dfrac 1{ 1+ \sec x} \right)\\\\\\=3 \dfrac{d}{dx} (1+ \sec x)^{-1}\\\\\\=3 (-1) (1 + \sec x )^{-1 -1} \dfrac{d}{dx}( 1 + \sec x)\\\\\\=-3(1 + \sec x)^{-2} ( 0 + \sec x \tan x)\\\\\\=-\dfrac{3\sec x \tan x}{(1 + \sec x)^2}$$
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