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Lt y tends to 0 [(x+y) sec(x+y) -x.secx] /y

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$\lim_{y \to 0} \frac{(x + y) \sec (x + y) - x \sec x}{y} = \lim_{y \to 0} \frac{x \sec (x + y) + y \sec (x + y) - x \sec x}{y} = \lim_{y \to 0} \frac{x [\sec (x + y) - \sec x] + y \sec (x + y)}{y} = \lim_{y \to 0} \frac{x [\sec (x + y) - \sec x]}{y} + \lim_{y \to 0} \frac{y \sec (x + y)}{y} = \lim_{y \to 0} \frac{x [\cos x - \cos (x + y)]}{y \cos x \cos (x + y)} + \lim_{y \to 0} \sec (x + y) = \lim_{y \to 0} [\frac{\cos x - \cos (x + y)}{y} . \frac{x}{\cos x \cos (x + y)}] + \sec x = \lim_{y \to 0} [\frac{2 \sin (x + \frac{y}{2}) \sin (\frac{y}{2})}{2 (\frac{y}{2})} . \frac{x}{\cos x \cos (x + y)}] + \sec x = \lim_{y \to 0} \sin (x + \frac{y}{2}) . \lim_{y \to 0} \frac{2 \sin (\frac{y}{2})}{2 (\frac{y}{2})} . \lim_{y \to 0} \frac{x}{\cos x \cos (x + y)} + \sec x = \sin x . 1 . \frac{x}{\cos^{2} x} + \sec x = x \tan x \sec x + \sec x$

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