Find derivative by first principle of xcosec(x)???I am confused if wse will take the x multiplied also as (x+h) or not Also e^rootx find by first principle

To find the derivative of the function \( f(x) = x \cdot \csc(x) \) using the first principle of derivatives, we will apply the definition of the derivative. The first principle states that the derivative of a function at a point is the limit of the average rate of change of the function as the interval approaches zero. Mathematically, this is expressed as:

f'(x) = lim (h → 0) [(f(x + h) - f(x)) / h]

Step-by-Step Derivation of f(x) = x * csc(x)

Let's break this down into manageable steps:

1. Define the Function

We start with:

f(x) = x \cdot \csc(x) = x / \sin(x)

2. Calculate f(x + h)

Next, we need to find \( f(x + h) \):

f(x + h) = (x + h) \cdot \csc(x + h) = (x + h) / \sin(x + h)

3. Set Up the Difference Quotient

Now, we can substitute these into the limit definition:

f'(x) = lim (h → 0) [((x + h) / \sin(x + h) - x / \sin(x)) / h]

4. Simplify the Expression

To simplify this, we need a common denominator:

= lim (h → 0) [( (x + h) \cdot \sin(x) - x \cdot \sin(x + h) ) / (h \cdot \sin(x) \cdot \sin(x + h))]

5. Expand and Simplify Further

Using the sine addition formula and properties of limits, we can further simplify the numerator:

= lim (h → 0) [(x \cdot \sin(x) + h \cdot \sin(x) - x \cdot (\sin(x) + h \cdot \cos(x))) / (h \cdot \sin(x) \cdot \sin(x + h))]

= lim (h → 0) [(h \cdot \sin(x) - h \cdot x \cdot \cos(x)) / (h \cdot \sin(x) \cdot \sin(x + h))]

6. Cancel h and Take the Limit

Now we can cancel \( h \) from the numerator and denominator:

= lim (h → 0) [( \sin(x) - x \cdot \cos(x)) / (\sin(x) \cdot \sin(x + h))]

As \( h \) approaches 0, \( \sin(x + h) \) approaches \( \sin(x) \), so we have:

= ( \sin(x) - x \cdot \cos(x)) / (\sin^2(x))

7. Final Result

Thus, the derivative of \( f(x) = x \cdot \csc(x) \) is:

f'(x) = \frac{\sin(x) - x \cdot \cos(x)}{\sin^2(x)}

Finding the Derivative of e^√x

Now, let's tackle the second part of your question regarding \( g(x) = e^{\sqrt{x}} \) using the first principle.

1. Define the Function

We start with:

g(x) = e^{\sqrt{x}}

2. Calculate g(x + h)

Next, we find \( g(x + h) \):

g(x + h) = e^{\sqrt{x + h}}

3. Set Up the Difference Quotient

Using the definition of the derivative:

g'(x) = lim (h → 0) [(e^{\sqrt{x + h}} - e^{\sqrt{x}}) / h]

4. Factor Out e^{√x}

We can factor out \( e^{\sqrt{x}} \):

= e^{\sqrt{x}} \cdot lim (h → 0) [(e^{\sqrt{x + h} - \sqrt{x}} - 1) / h]

5. Use the Exponential Limit

As \( h \) approaches 0, \( \sqrt{x + h} - \sqrt{x} \) can be approximated using the derivative of \( \sqrt{x} \):

= \frac{1}{2\sqrt{x}} \cdot h

Thus, we can rewrite the limit as:

= e^{\sqrt{x}} \cdot lim (h → 0) [(e^{\frac{1}{2\sqrt{x}} \cdot h} - 1) / h]

6. Evaluate the Limit

Using the fact that \( lim (u → 0) \frac{e^u - 1}{u} = 1 \), we find:

= e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}

7. Final Result

Therefore, the derivative of \( g(x) = e^{\sqrt{x}} \) is:

g'(x) = \frac{e^{\sqrt{x}}}{2\sqrt{x}}

In summary, both derivatives were calculated using the first principle, demonstrating the application of limits and algebraic manipulation. If you have further questions or need clarification on any