question 16
find $\dfrac{dy}{dx}$ for $y = \sec x$.
$\circ$ $\dfrac{dy}{dx} = \sec^2 x$
$\circ$ $\dfrac{dy}{dx} = -\csc^2 x$
$\circ$ $\dfrac{dy}{dx} = -\csc x \cot x$
$\circ$ no correct answer choice is given.
$\circ$ $\dfrac{dy}{dx} = \sec x \tan x$

Question

Accepted Answer

# Explanation:
## Step1: Recall the derivative of secant function
The derivative of $ y = \sec x $ can be found using the definition of secant as $ \sec x=\frac{1}{\cos x} $. We use the quotient rule, which states that if $ y = \frac{u}{v} $, then $ y'=\frac{u'v - uv'}{v^{2}} $. Here, $ u = 1 $ and $ v=\cos x $.
## Step2: Compute the derivatives of u and v
The derivative of $ u = 1 $ with respect to $ x $ is $ u'=0 $, and the derivative of $ v=\cos x $ with respect to $ x $ is $ v'=-\sin x $.
## Step3: Apply the quotient rule
Substitute $ u, v, u', v' $ into the quotient rule:
$$
\begin{align*}
\frac{dy}{dx}&=\frac{0\times\cos x-1\times(-\sin x)}{\cos^{2}x}\
&=\frac{\sin x}{\cos^{2}x}\
&=\frac{1}{\cos x}\times\frac{\sin x}{\cos x}\
&=\sec x\tan x
\end{align*}
$$

# Answer:
$ \boldsymbol{\frac{dy}{dx}=\sec x\tan x} $ (the last option: $ \frac{dy}{dx}=\sec x\tan x $)

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the derivative of secant function

Step2: Compute the derivatives of u and v

Step3: Apply the quotient rule

Answer: