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

Answer

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

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

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

### Step 3: Factor out common terms.
We can factor out \( e^{\sin x} \) from the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{e^{\sin x} \left( e^{\sin(x+h) - \sin x} - 1 \right)}{h} \]

### Step 4: Simplify the expression involving \( \sin \).
Now, we need to analyze \( \sin(x+h) - \sin x \):
Using the sine addition formula:
\[ \sin(x+h) = \sin x \cos h + \cos x \sin h \]
Thus,
\[ \sin(x+h) - \sin x = \sin x \cos h + \cos x \sin h - \sin x = \cos x \sin h + \sin x (\cos h - 1) \]

### Step 5: Substitute back into the limit.
Now we substitute this back into our limit:
\[ f'(x) = \lim_{h \to 0} \frac{e^{\sin x} \left( e^{\cos x \sin h + \sin x (\cos h - 1)} - 1 \right)}{h} \]

### Step 6: Use the limit property for small angles.
As \( h \to 0 \), \( \sin h \to h \) and \( \cos h \to 1 \). Therefore, we can use the property:
\[ \lim_{u \to 0} \frac{e^u - 1}{u} = 1 \]
Let \( u = \cos x \sin h + \sin x (\cos h - 1) \). As \( h \to 0 \), this approaches \( \cos x \cdot h \).

### Step 7: Evaluate the limit.
Thus, we have:
\[ f'(x) = e^{\sin x} \cdot \lim_{h \to 0} \frac{\cos x \sin h + \sin x (\cos h - 1)}{h} \]
This simplifies to:
\[ f'(x) = e^{\sin x} \cdot \left( \cos x \cdot 1 + \sin x \cdot 0 \right) = e^{\sin x} \cdot \cos x \]

### Final Answer:
Thus, the derivative of \( f(x) = e^{\sin x} \) is:
\[ f'(x) = e^{\sin x} \cos x \]

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