$\begin{align}  & y=x\sec x \\ & \text{Let }x\text{ be changed to }x+\delta x\text{ and consequently }y\text{ to }y+\delta y \\ & y+\delta y=(x+\delta x)\sec (x+\delta x) \\ & \delta y=(x+\delta x)\sec (x+\delta x)-x\sec x \\ & \delta y=\frac{x+\delta x}{\cos (x+\delta x)}-\frac{x}{\cos x}=\frac{(x+\delta x)\cos x-x\cos (x+\delta x)}{\cos x\cos (x+\delta x)} \\ & \delta y=\frac{x\cos x+\cos x\delta x-x\cos (x+\delta x)}{\cos x\cos (x+\delta x)} \\ & \delta y=\frac{x\cos x-x\cos (x+\delta x)}{\cos x\cos (x+\delta x)}+\frac{\cos x\delta x}{\cos x\cos (x+\delta x)} \\ & \delta y=\frac{x\left[ 2\sin \tfrac{x+x+\delta x}{2}\sin \tfrac{x+\delta x-x}{2} \right]}{\cos x\cos (x+\delta x)}+\frac{\cos x\delta x}{\cos x\cos (x+\delta x)} \\ & \text{Recall that }\cos A-\cos B=2\sin \tfrac{A+B}{2}\sin \tfrac{B-A}{2} \\ & \delta y=\frac{2x\sin (x+\tfrac{\delta x}{x})\sin \tfrac{\delta x}{2}}{\cos x\cos (x+\delta x)}+\frac{\delta x}{\cos (x+\delta x)} \\ & \text{Divide through by }\delta x \\ & \frac{\delta y}{\delta x}=\frac{1}{\delta x}\left[ \frac{2x\sin (x+\tfrac{\delta x}{x})\sin \tfrac{\delta x}{2}}{\cos x\cos (x+\delta x)}+\frac{\delta x}{\cos (x+\delta x)} \right] \\ & \frac{\delta y}{\delta x}=\frac{x\sin (x+\delta x)}{\cos x\cos (x+\delta x)}\cdot \frac{\sin \tfrac{\delta x}{2}}{\tfrac{\delta x}{2}}+\frac{1}{\cos (x+\delta x)} \\ & \text{proceeding to the limits as }\delta x\to 0 \\ & \frac{dy}{dx}=\underset{\delta x\to 0}{\mathop{\lim }}\,\frac{\delta y}{\delta x}=x\underset{\delta x\to 0}{\mathop{\lim }}\,\frac{x\sin (x+\delta x)}{\cos x\cos (x+\delta x)}\underset{\delta x\to 0}{\mathop{\lim }}\,\frac{\sin \tfrac{\delta x}{2}}{\tfrac{\delta x}{2}}+\underset{\delta x\to 0}{\mathop{\lim }}\,\frac{1}{\cos (x+\delta x)} \\ & \frac{dy}{dx}=x\frac{\sin x}{{{\cos }^{2}}x}+\frac{1}{\cos x}=x\frac{\sin x}{\cos x\cos x}+\frac{1}{\cos x} \\ & \frac{dy}{dx}=x\tan x\sec x+\sec x \\\end{align}$