Mathematics

Problem

The answers are in Chapter 24. 5. Find

\frac{d y}{d x}

y=\sqrt{\frac{x^{2}}{\sec x}}

. (A)

\frac{d y}{d x}=\frac{1}{2}\left(\frac{2 x}{\sec x \tan x}\right)^{-\frac{1}{2}}

(B)

\frac{d y}{d x}=\left(\frac{2 x}{\sec x \tan x}\right)^{\frac{1}{2}}

(C)

\frac{d y}{d x}=\frac{1}{2}\left(\frac{\sec x(2 x)-\left(x^{2}\right) \sec x \tan x}{\sec ^{2} x}\right)^{-\frac{1}{2}}

(D)

\frac{d y}{d x}=\frac{1}{2}\left(\frac{x^{2}}{\sec x}\right)^{-\frac{1}{2}}\left(\frac{\sec x(2 x)-\left(x^{2}\right) \sec x \tan x}{\sec ^{2} x}\right)

(A)

\frac{d y}{d x}=\frac{1}{2}\left(\frac{2 x}{\sec x \tan x}\right)^{-\frac{1}{2}}

(B)

\frac{d y}{d x}=\left(\frac{2 x}{\sec x \tan x}\right)^{\frac{1}{2}}

(C)

\frac{d y}{d x}=\frac{1}{2}\left(\frac{\sec x(2 x)-\left(x^{2}\right) \sec x \tan x}{\sec ^{2} x}\right)^{-\frac{1}{2}}

(D)

\frac{d y}{d x}=\frac{1}{2}\left(\frac{x^{2}}{\sec x}\right)^{-\frac{1}{2}}\left(\frac{\sec x(2 x)-\left(x^{2}\right) \sec x \tan x}{\sec ^{2} x}\right)