Differentiation: Question 6

Worked solution

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

$y = 5x^4 - 2x^{-3} + 6x^{1/3}$

Writing the negative and fractional 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: $5x^4 \to 5 \times 4\, x^{4-1} = 20x^3$

Term 2: $-2x^{-3} \to -2 \times (-3)\, x^{-3-1} = 6x^{-4} = \dfrac{6}{x^4}$

(Two negatives, the $-2$ and the $-3$, multiply to give a positive $6$.)

Term 3: $6x^{1/3} \to 6 \times \tfrac13\, x^{1/3-1} = 2x^{-2/3} = \dfrac{2}{\sqrt[3]{x^2}}$

Adding these together:

$\frac{dy}{dx} = 20x^3 + \frac{6}{x^4} + \frac{2}{\sqrt[3]{x^2}}$

Step 3: Recompute independently as a check

Differentiating again from scratch, term by term:

• $\dfrac{d}{dx}(5x^4)$: bring down the power $4$ as a coefficient, reduce the power by $1$: $20x^3$. ✓
• $\dfrac{d}{dx}(-2x^{-3})$: bring down the power $-3$ as a coefficient: $(-2)(-3)x^{-4} = 6x^{-4}$. ✓
• $\dfrac{d}{dx}(6x^{1/3})$: bring down the power $\tfrac13$ as a coefficient: $6 \times \tfrac13\, x^{-2/3} = 2x^{-2/3}$. ✓

This matches Step 2 exactly, so

$\frac{dy}{dx} = 20x^3 + \frac{6}{x^4} + \frac{2}{\sqrt[3]{x^2}}.$

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 $(-2)\times(-3) = +6$, not $-6$.
• B: the first term’s power was not reduced from $4$ to $3$ when differentiating $5x^4$.
• D: the coefficient of the last term was not multiplied by $\tfrac13$, so $6$ was carried over unchanged instead of becoming $2$.

Final answer

$\boxed{\frac{dy}{dx} = 20x^3 + \frac{6}{x^4} + \frac{2}{\sqrt[3]{x^2}}} \quad \text{(Option C)}$