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 need to find the value of \(\frac{1}{4} f'(27)\) given that \(f(x^3) = x^5\) for all \(x \in \mathbb{R} - \{0\}\) and that \(f(x)\) is differentiable for all \(x \in \mathbb{R}\).

### Step-by-Step Solution:

1. **Understanding the function:**
We have the equation \(f(x^3) = x^5\). We need to express \(f(x)\) in terms of \(x\).

2. **Substituting \(y = x^3\):**
Let \(y = x^3\). Then, we can express \(x\) in terms of \(y\):
\[ x = y^{1/3} \]
Substituting this into the equation for \(f\):
\[ f(y) = (y^{1/3})^5 = y^{5/3} \]
Thus, we have:
\[ f(x) = x^{5/3} \quad \text{for } x \in \mathbb{R} - \{0\} \]

3. **Finding the derivative \(f'(x)\):**
Now we differentiate \(f(x)\):
\[ f'(x) = \frac{d}{dx}(x^{5/3}) = \frac{5}{3} x^{(5/3) - 1} = \frac{5}{3} x^{2/3} \]

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

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

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