Differentiate the following (7-10) with respect to 'x' using first principle :
`x^(2/3)`

Answer

To differentiate \( y = x^{2/3} \) with respect to \( x \) using the first principle of derivatives, we follow these steps:

### Step 1: Set up the definition of the derivative
The derivative of a function \( y = f(x) \) using the first principle is defined as:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute the function into the definition
For our function \( f(x) = x^{2/3} \), we need to calculate \( f(x+h) \):
\[ f(x+h) = (x+h)^{2/3} \]
Thus, we can write:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{(x+h)^{2/3} - x^{2/3}}{h} \]

### Step 3: Simplify the expression
We can use the identity for the difference of cubes:
\[ a^2 - b^2 = (a-b)(a+b) \]
In this case, we can rewrite \( (x+h)^{2/3} - x^{2/3} \) as:
\[ \frac{(x+h)^{2/3} - x^{2/3}}{h} = \frac{(x+h)^{2/3} - x^{2/3}}{(x+h)^{1/3} + x^{1/3}} \cdot \frac{(x+h)^{1/3} + x^{1/3}}{h} \]

### Step 4: Apply the limit
Now, we can apply the limit:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{(x+h)^{2/3} - x^{2/3}}{(x+h)^{1/3} + x^{1/3}} \cdot \frac{1}{h} \]
As \( h \to 0 \), \( (x+h)^{2/3} \to x^{2/3} \) and \( (x+h)^{1/3} \to x^{1/3} \).

### Step 5: Evaluate the limit
Now, substituting \( h = 0 \):
\[ \frac{dy}{dx} = \frac{2}{3} x^{-1/3} \]

### Final Result
Thus, the derivative of \( y = x^{2/3} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{2}{3} x^{-1/3} \]