Answer
$$f'(x)=\frac{\sec x(1+x\tan x)}{2\sqrt{7+x\sec x}}$$
Work Step by Step
$$f(x)=\sqrt{7+x\sec x}=(7+x\sec x)^{1/2}$$
The derivative of function $f(x)$ is $$f'(x)=\frac{1}{2}(7+x\sec x)^{-1/2}(7+x\sec x)'$$
$$f'(x)=\frac{1}{2}(7+x\sec x)^{-1/2}\Big(0+(x)'\sec x+x(\sec x)'\Big)$$
$$f'(x)=\frac{1}{2\sqrt{7+x\sec x}}(\sec x+x\sec x\tan x)$$
$$f'(x)=\frac{\sec x(1+x\tan x)}{2\sqrt{7+x\sec x}}$$
