Let's break down and solve both questions from the image.

Question 13:

The figure shows the function $f(x)$, and the straight line $L$ touches the curve at the point $A(3, 5)$. The function $g(x)$ is given as: $g(x) = x \cdot f(x)$ We are asked to find $g'(3)$, the derivative of $g(x)$ at $x = 3$.

Approach:

Using the product rule to differentiate $g(x) = x \cdot f(x)$, we have: $g'(x) = \frac{d}{dx}[x \cdot f(x)] = f(x) + x \cdot f'(x)$ At $x = 3$: $g'(3) = f(3) + 3 \cdot f'(3)$ From the information given, $f(3) = 5$, and since the line $L$ is tangent to the curve at $A(3, 5)$, the slope of the tangent line $f'(3)$ is the slope of the line. From the diagram, the slope of $L$ appears to be $\tan(45^\circ) = 1$. Therefore: $g'(3) = 5 + 3 \cdot 1 = 5 + 3 = 8$ Thus, the answer is 8.

Answer: (d) 8

Question 14:

The given function is: $y = \sqrt{2 (\sec x + \tan x)}$ We need to find $\frac{1}{y} \cdot \frac{dy}{dx}$.

Approach:

1. First, differentiate $y$ with respect to $x$. Let: $y = \left[ 2 (\sec x + \tan x) \right]^{1/2}$ Using the chain rule, we first differentiate the square root function, and then the inner function: [ \frac{dy}{dx} = \frac{1}{2} \left[ 2 (\sec x + \tan x) \right]^{-1/2} \cdot \frac{d}{dx} \left[ 2 (\sec x + \tan x) \right] ]
2. The derivative of $2 (\sec x + \tan x)$ is: $\frac{d}{dx} \left[ 2 (\sec x + \tan x) \right] = 2 (\sec x \tan x + \sec^2 x)$ So, the derivative becomes: $\frac{dy}{dx} = \frac{1}{2} \cdot \frac{2 (\sec x \tan x + \sec^2 x)}{\sqrt{2 (\sec x + \tan x)}}$ Simplifying: $\frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{\sqrt{2 (\sec x + \tan x)}}$
3. Now, we calculate $\frac{1}{y} \cdot \frac{dy}{dx}$: $\frac{1}{y} = \frac{1}{\sqrt{2 (\sec x + \tan x)}}$ Therefore: $\frac{1}{y} \cdot \frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{2 (\sec x + \tan x)}$
4. Simplifying this expression: $\frac{1}{y} \cdot \frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{2 (\sec x + \tan x)} = \frac{\sec x}{2}$ Thus, the answer is: $\boxed{\frac{1}{2} \sec x}$

Answer: (a) $\frac{1}{2} \sec x$

Let me know if you need more details on any of these, or if you'd like clarification on the steps.

5 Related Questions:

1. What is the product rule for differentiation and how is it used in calculus?
2. How can we find the slope of a tangent line to a curve at a given point?
3. What is the derivative of trigonometric functions like $\sec x$ and $\tan x$?
4. How do we apply the chain rule when differentiating composite functions?
5. What is the significance of the derivative in terms of the rate of change and slopes?

Tip:

When dealing with composite functions or products of functions, always remember to apply the chain rule or product rule, respectively, to ensure the correct differentiation.