Math / Calculus

Question(1) If $x=\sec (z), \sqrt{y}=\tan (z)$ , find: $\frac{d^{2} y}{d x^{2}}$ (A) 0 (B) 1 (C) 2 (D) 3

Studdy Solution

STEP 1

1. The given problem involves trigonometric functions and their relationships with derivatives.
2. We need to find the second derivative of $y$ with respect to $x$ , where $x = \sec(z)$ and $\sqrt{y} = \tan(z)$ .
3. Chain rule and implicit differentiation will be essential tools in this problem.

STEP 2

1. Express $y$ in terms of $z$ .
2. Differentiate $y$ with respect to $z$ .
3. Relate $dx$ and $dz$ using implicit differentiation.
4. Use chain rule to find the first derivative of $y$ with respect to $x$ .
5. Differentiate the first derivative to find the second derivative of $y$ with respect to $x$ .

STEP 3

Express $y$ in terms of $z$ .
Given $\sqrt{y} = \tan(z)$ , square both sides to get: $y = \tan^2(z)$

STEP 4

Differentiate $y$ with respect to $z$ .
Using the chain rule: $\frac{dy}{dz} = \frac{d}{dz}(\tan^2(z)) = 2\tan(z) \cdot \sec^2(z)$

STEP 5

Relate $dx$ and $dz$ using implicit differentiation.
Given $x = \sec(z)$ , differentiate both sides with respect to $z$ : $\frac{dx}{dz} = \sec(z) \tan(z)$

STEP 6

Use the chain rule to find the first derivative of $y$ with respect to $x$ .
From the chain rule: $\frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx}$
We already have $\frac{dy}{dz} = 2 \tan(z) \sec^2(z)$ and $\frac{dx}{dz} = \sec(z) \tan(z)$ .
So, $\frac{dz}{dx} = \frac{1}{\frac{dx}{dz}} = \frac{1}{\sec(z) \tan(z)}$
Therefore, $\frac{dy}{dx} = 2 \tan(z) \sec^2(z) \cdot \frac{1}{\sec(z) \tan(z)} = 2 \sec(z)$

STEP 7

Differentiate the first derivative to find the second derivative of $y$ with respect to $x$ .
$\frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} (2 \sec(z))$
Using the chain rule again: $\frac{d}{dx} (2 \sec(z)) = 2 \frac{d}{dz}(\sec(z)) \cdot \frac{dz}{dx}$
Since $\frac{d}{dz}(\sec(z)) = \sec(z) \tan(z)$ and $\frac{dz}{dx} = \frac{1}{\sec(z) \tan(z)}$ ,
$\frac{d}{dx} (2 \sec(z)) = 2 \sec(z) \tan(z) \cdot \frac{1}{\sec(z) \tan(z)} = 2$
So, the second derivative is: $\frac{d^2 y}{d x^2} = 2$

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