If `f:Rrarr R` is a function defined as `f(x^(3))=x^(5), AA x in R -{0} and f(x)` is differentiable `AA x in R,` then the value of `(1)/(4)f'(27)` is equal to (here f' represents the derivative of f)

Answer

To solve the problem, we start with the given function \( f \) defined by the equation \( f(x^3) = x^5 \) for \( x \in \mathbb{R} - \{0\} \). We need to find the value of \( \frac{1}{4} f'(27) \).

### Step-by-step Solution:

1. **Understanding the Function**:
We have the function defined as \( f(x^3) = x^5 \). We can express \( f(y) \) in terms of \( y \) by letting \( y = x^3 \). Thus, we can write:
\[ x = y^{1/3} \implies f(y) = (y^{1/3})^5 = y^{5/3} \]
Therefore, we have:
\[ f(y) = y^{5/3} \]

2. **Finding the Derivative**:
Next, we differentiate \( f(y) \):
\[ f'(y) = \frac{d}{dy}(y^{5/3}) = \frac{5}{3}y^{2/3} \]

3. **Calculating \( f'(27) \)**:
Now we need to find \( f'(27) \):
\[ f'(27) = \frac{5}{3} \cdot 27^{2/3} \]
To compute \( 27^{2/3} \):
\[ 27^{2/3} = (3^3)^{2/3} = 3^2 = 9 \]
Therefore,
\[ f'(27) = \frac{5}{3} \cdot 9 = 15 \]

4. **Finding \( \frac{1}{4} f'(27) \)**:
Finally, we compute \( \frac{1}{4} f'(27) \):
\[ \frac{1}{4} f'(27) = \frac{1}{4} \cdot 15 = \frac{15}{4} = 3.75 \]

Thus, the value of \( \frac{1}{4} f'(27) \) is \( 3.75 \).