We need to find the value of the derivative $\frac{\mathrm{d} }{\mathrm{d} x}\left [ \frac{f(x^3)}{x f(x^2)} \right ]$ at $x = 1$, given $f(1) = 1$ and $f'(1) = -1$. Let the function be $y = \frac{f(x^3)}{x f(x^2)}$.

Derivative Calculation using Quotient Rule

We use the quotient rule $\frac{\mathrm{d} }{\mathrm{d} x}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$.

Let $u = f(x^3)$ and $v = x f(x^2)$.

Derivative of Numerator ($u'$)

Using the chain rule, $\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} f(x^3) = f'(x^3) \cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^3) = f'(x^3) \cdot 3x^2$.

Derivative of Denominator ($v'$)

Using the product rule and chain rule, $\frac{\mathrm{d} v}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} [x f(x^2)] = (1) \cdot f(x^2) + x \cdot \frac{\mathrm{d} }{\mathrm{d} x}[f(x^2)]$.

Applying the chain rule again: $\frac{\mathrm{d} }{\mathrm{d} x}[f(x^2)] = f'(x^2) \cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2) = f'(x^2) \cdot 2x$.

So, $v' = f(x^2) + x [f'(x^2) \cdot 2x] = f(x^2) + 2x^2 f'(x^2)$.

Full Derivative Expression

Substituting $u, v, u', v'$ into the quotient rule:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{(f'(x^3) \cdot 3x^2) (x f(x^2)) - (f(x^3)) (f(x^2) + 2x^2 f'(x^2))}{(x f(x^2))^2}$

Evaluation at x = 1

Now, we evaluate the derivative at $x = 1$, using the given values $f(1) = 1$ and $f'(1) = -1$.

• $u(1) = f(1^3) = f(1) = 1$.
• $v(1) = 1 \cdot f(1^2) = 1 \cdot f(1) = 1$.
• $u'(1) = f'(1^3) \cdot 3(1)^2 = f'(1) \cdot 3 = (-1) \cdot 3 = -3$.
• $v'(1) = f(1^2) + 2(1)^2 f'(1^2) = f(1) + 2 f'(1) = 1 + 2(-1) = 1 - 2 = -1$.

Substitute these values into the derivative formula:

$\frac{\mathrm{d} y}{\mathrm{d} x}\bigg|_{x=1} = \frac{u'(1)v(1) - u(1)v'(1)}{[v(1)]^2} = \frac{(-3)(1) - (1)(-1)}{(1)^2}$

$\frac{\mathrm{d} y}{\mathrm{d} x}\bigg|_{x=1} = \frac{-3 - (-1)}{1} = \frac{-3 + 1}{1} = -2$.