Differentiation: Question 1

Worked solution

Step 1: Rewrite every term as a power of $x$

$y = 4x^3 - 6x^{-2} + 8x^{1/2}$

Writing the fractional and negative powers explicitly makes it safe to apply the power rule $\dfrac{d}{dx}(x^n) = nx^{n-1}$ to each term separately.

Step 2: Differentiate term by term

Term 1: $4x^3 \to 4 \times 3\, x^{3-1} = 12x^2$

Term 2: $-6x^{-2} \to -6 \times (-2)\, x^{-2-1} = 12x^{-3} = \dfrac{12}{x^3}$

(Two negatives, the $-6$ and the $-2$, multiply to give a positive $12$.)

Term 3: $8x^{1/2} \to 8 \times \tfrac12\, x^{1/2 - 1} = 4x^{-1/2} = \dfrac{4}{\sqrt{x}}$

Adding these together:

$\frac{dy}{dx} = 12x^2 + \frac{12}{x^3} + \frac{4}{\sqrt{x}}$

Step 3: Recompute independently as a check

Differentiating again from scratch, term by term:

• $\dfrac{d}{dx}(4x^3)$: bring down the power $3$ as a coefficient, reduce the power by $1$: $12x^2$. ✓
• $\dfrac{d}{dx}(-6x^{-2})$: bring down the power $-2$ as a coefficient: $(-6)(-2)x^{-3} = 12x^{-3}$. ✓
• $\dfrac{d}{dx}(8x^{1/2})$: bring down the power $\tfrac12$ as a coefficient: $8 \times \tfrac12 \, x^{-1/2} = 4x^{-1/2}$. ✓

This matches Step 2 exactly, so

$\frac{dy}{dx} = 12x^2 + \frac{12}{x^3} + \frac{4}{\sqrt{x}}.$

Why the other options are wrong

• A: has the correct size for every term, but the sign of the middle term is wrong. It comes from forgetting that $(-6)\times(-2) = +12$, not $-12$.
• B: the first term’s power was not reduced from $3$ to $2$ when differentiating $4x^3$.
• D: the coefficient of the last term was not multiplied by $\tfrac12$, so $8$ was carried over unchanged instead of becoming $4$.

Final answer

$\boxed{\frac{dy}{dx} = 12x^2 + \frac{12}{x^3} + \frac{4}{\sqrt{x}}} \quad \text{(Option C)}$