Find, from first principle, the derivative of the following w.r.t. x :
`e^(sqrttanx)`

Answer

To find the derivative of the function \( f(x) = e^{\sqrt{\tan x}} \) using the first principle of derivatives, we follow these steps:

### Step 1: Write the definition of the derivative
The derivative of a function \( f(x) \) from first principles is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute the function into the definition
Substituting \( f(x) = e^{\sqrt{\tan x}} \) into the definition, we have:
\[ f'(x) = \lim_{h \to 0} \frac{e^{\sqrt{\tan(x+h)}} - e^{\sqrt{\tan x}}}{h} \]

### Step 3: Factor out the common term
We can factor out \( e^{\sqrt{\tan x}} \) from the numerator:
\[ f'(x) = e^{\sqrt{\tan x}} \lim_{h \to 0} \frac{e^{\sqrt{\tan(x+h)}} - e^{\sqrt{\tan x}}}{h} \]

### Step 4: Use the exponential limit property
Using the property of limits, we can rewrite the limit:
\[ \lim_{h \to 0} \frac{e^{u} - e^{v}}{h} = e^{v} \cdot \lim_{h \to 0} \frac{e^{u} - e^{v}}{u - v} \cdot \frac{u - v}{h} \]
where \( u = \sqrt{\tan(x+h)} \) and \( v = \sqrt{\tan x} \).

### Step 5: Simplify the limit
Now we need to find \( u - v \):
\[ u - v = \sqrt{\tan(x+h)} - \sqrt{\tan x} \]
We can multiply and divide by the conjugate:
\[ u - v = \frac{\tan(x+h) - \tan x}{\sqrt{\tan(x+h)} + \sqrt{\tan x}} \]

### Step 6: Substitute back into the limit
Substituting back, we have:
\[ f'(x) = e^{\sqrt{\tan x}} \cdot \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h(\sqrt{\tan(x+h)} + \sqrt{\tan x})} \]

### Step 7: Use the derivative of tangent
The derivative of \( \tan x \) is \( \sec^2 x \), so:
\[ \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h} = \sec^2 x \]

### Step 8: Combine everything
Putting it all together, we get:
\[ f'(x) = e^{\sqrt{\tan x}} \cdot \sec^2 x \cdot \frac{1}{2\sqrt{\tan x}} \cdot \lim_{h \to 0} \frac{1}{\sqrt{\tan(x+h)} + \sqrt{\tan x}} \]

As \( h \to 0 \), \( \sqrt{\tan(x+h)} \to \sqrt{\tan x} \), thus:
\[ \lim_{h \to 0} \frac{1}{\sqrt{\tan(x+h)} + \sqrt{\tan x}} = \frac{1}{2\sqrt{\tan x}} \]

### Final Result
Combining all parts, we find:
\[ f'(x) = e^{\sqrt{\tan x}} \cdot \sec^2 x \cdot \frac{1}{2\sqrt{\tan x}} \cdot \frac{1}{2\sqrt{\tan x}} = \frac{e^{\sqrt{\tan x}} \sec^2 x}{2 \tan x} \]

Thus, the derivative of \( e^{\sqrt{\tan x}} \) is:
\[ f'(x) = e^{\sqrt{\tan x}} \cdot \sec^2 x \cdot \frac{1}{2\sqrt{\tan x}} \]