Differentiate the following with respect to x using first principle method.
`sin^((1)/(3))x=root(3)(sinx)`.

Answer

To differentiate the function \( f(x) = \sqrt[3]{\sin x} \) using the first principle method, we will follow these steps:

### Step 1: Define the function
We start with the function:
\[ f(x) = \sqrt[3]{\sin x} \]

### Step 2: Apply the first principle of differentiation
According to the first principle of differentiation, the derivative \( f'(x) \) is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Substituting our function into this formula gives:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt[3]{\sin(x+h)} - \sqrt[3]{\sin x}}{h} \]

### Step 3: Use the identity for the difference of cubes
We can use the identity for the difference of cubes:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Let \( a = \sqrt[3]{\sin(x+h)} \) and \( b = \sqrt[3]{\sin x} \). Then:
\[ \sqrt[3]{\sin(x+h)} - \sqrt[3]{\sin x} = \frac{\sin(x+h) - \sin x}{\sqrt[3]{\sin^2(x+h)} + \sqrt[3]{\sin(x+h)\sin x} + \sqrt[3]{\sin^2 x}} \]

### Step 4: Substitute back into the limit
Substituting this back into our limit gives:
\[ f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h \left( \sqrt[3]{\sin^2(x+h)} + \sqrt[3]{\sin(x+h)\sin x} + \sqrt[3]{\sin^2 x} \right)} \]

### Step 5: Simplify using the derivative of sine
Using the derivative of sine, we know:
\[ \sin(x+h) - \sin x = 2 \cos\left(\frac{x+h+x}{2}\right) \sin\left(\frac{h}{2}\right) \]
Thus, we can rewrite our limit:
\[ f'(x) = \lim_{h \to 0} \frac{2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h \left( \sqrt[3]{\sin^2(x+h)} + \sqrt[3]{\sin(x+h)\sin x} + \sqrt[3]{\sin^2 x} \right)} \]

### Step 6: Evaluate the limit
As \( h \to 0 \), \( \sin\left(\frac{h}{2}\right) \approx \frac{h}{2} \). Thus:
\[ f'(x) = \lim_{h \to 0} \frac{2 \cos(x) \cdot \frac{h}{2}}{h \left( \sqrt[3]{\sin^2 x} + \sqrt[3]{\sin^2 x} + \sqrt[3]{\sin^2 x} \right)} = \frac{\cos x}{3 \sqrt[3]{\sin^2 x}} \]

### Final Result
Thus, the derivative of \( f(x) = \sqrt[3]{\sin x} \) is:
\[ f'(x) = \frac{\cos x}{3 \sqrt[3]{\sin^2 x}} \]