14) If $$ y=\sin x+\sec x, x=3 \pi z $$, then $$ \frac{d y}{d z}=\ldots \ldots \ldots $$. At $$ z=1 $$ $$ \begin{array}{llll} ext { a) } 3 \pi & ext { b) }-3 \pi & ext { c) }-6 \pi & ext { d) }-1\end{array} $$ b

Question

14) If $$ y=\sin x+\sec x, x=3 \pi z $$, then $$ \frac{d y}{d z}=\ldots \ldots \ldots $$. At $$ z=1 $$ $$ \begin{array}{llll}	ext { a) } 3 \pi & 	ext { b) }-3 \pi & 	ext { c) }-6 \pi & 	ext { d) }-1\end{array} $$ b

UpStudy AI · Accepted Answer

Let's verify the derivative step by step to ensure that option **b) $$-3\pi$$** is indeed the correct answer.

### Given:
$$ y = \sin x + \sec x $$
$$ x = 3\pi z $$

We need to find:
$$ \frac{dy}{dz} \quad \text{at} \quad z = 1 $$

### Step 1: Find $$\frac{dx}{dz}$$
Since $$ x = 3\pi z $$, the derivative of $$ x $$ with respect to $$ z $$ is:
$$ \frac{dx}{dz} = 3\pi $$

### Step 2: Find $$\frac{dy}{dx}$$
Differentiate $$ y $$ with respect to $$ x $$:
$$ \frac{dy}{dx} = \frac{d}{dx} (\sin x) + \frac{d}{dx} (\sec x) $$
$$ \frac{dy}{dx} = \cos x + \sec x \tan x $$

### Step 3: Apply the Chain Rule to Find $$\frac{dy}{dz}$$
Using the chain rule:
$$ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} $$
$$ \frac{dy}{dz} = (\cos x + \sec x \tan x) \cdot 3\pi $$

### Step 4: Evaluate at $$ z = 1 $$
First, find $$ x $$ when $$ z = 1 $$:
$$ x = 3\pi \times 1 = 3\pi $$

Now, evaluate each trigonometric function at $$ x = 3\pi $$:
- $$ \cos(3\pi) = \cos(\pi) = -1 $$ (since cosine has a period of $$ 2\pi $$)
- $$ \sec(3\pi) = \frac{1}{\cos(3\pi)} = \frac{1}{-1} = -1 $$
- $$ \tan(3\pi) = 0 $$ (since tangent of any multiple of $$ \pi $$ is 0)

Plug these values back into the derivative:
$$ \frac{dy}{dz} = (-1) + (-1)(0) $$
$$ \frac{dy}{dz} = -1 $$

Now multiply by $$ 3\pi $$:
$$ \frac{dy}{dz} = -1 \times 3\pi = -3\pi $$

### Conclusion:
The correct derivative at $$ z = 1 $$ is:
$$ \frac{dy}{dz} = -3\pi $$

So, **option b) $$-3\pi$$** is indeed the correct answer.
At $$ z = 1 $$, $$ \frac{dy}{dz} = -3\pi $$.

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b

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