Differentiation: Question 10

Worked solution

Part (a): Finding $\dfrac{dy}{dx}$

Rewrite the curve as a sum of powers of $x$:

$y = x^2 + 16x^{-1}$

Differentiating term by term:

$\frac{dy}{dx} = 2x + 16\times(-1)\,x^{-2} = 2x - 16x^{-2} = 2x - \frac{16}{x^2}$

Recomputing independently: $\dfrac{d}{dx}(x^2) = 2x$, and $\dfrac{d}{dx}(16x^{-1})$ brings down the power $-1$ as a coefficient, giving $16\times(-1)x^{-2}=-16x^{-2}$. Adding these confirms

$\frac{dy}{dx} = 2x - \frac{16}{x^2}.$

Part (b): Stationary point and its nature

Stationary points occur where $\dfrac{dy}{dx} = 0$:

$2x - \frac{16}{x^2} = 0$

Multiply through by $x^2$ (valid since $x \neq 0$):

$2x^3 - 16 = 0 \implies x^3 = 8 \implies x = 2$

(This is the only real cube root of $8$.)

Substitute back into $y = x^2 + \dfrac{16}{x}$:

$y = 2^2 + \frac{16}{2} = 4 + 8 = 12$

So the stationary point is $(2, 12)$.

Differentiate again to classify it:

$\frac{d^2y}{dx^2} = 2 + 32x^{-3} = 2 + \frac{32}{x^3}$

At $x = 2$:

$\frac{d^2y}{dx^2} = 2 + \frac{32}{8} = 2 + 4 = 6 > 0$

so $(2, 12)$ is a minimum point.

Independent check using a sign test either side of $x=2$. At $x=1$: $\dfrac{dy}{dx} = 2(1) - 16 = -14 < 0$ (curve decreasing just before $x=2$). At $x=3$: $\dfrac{dy}{dx} = 2(3) - \dfrac{16}{9} = 6 - 1.78\ldots = 4.22\ldots > 0$ (curve increasing just after $x=2$). The gradient changes from negative to positive at $x=2$, confirming a minimum, consistent with the second-derivative test.

Part (c): Tangent at $x = 4$

The $y$-coordinate at $x=4$:

$y = 4^2 + \frac{16}{4} = 16 + 4 = 20$

so the point is $(4, 20)$.

The gradient at $x=4$:

$\frac{dy}{dx}\bigg|_{x=4} = 2(4) - \frac{16}{4^2} = 8 - 1 = 7$

The tangent at $(4,20)$ with gradient $7$:

$y - 20 = 7(x - 4) \implies y = 7x - 28 + 20 = 7x - 8$

Check: substituting $x=4$ back in, $y = 7(4) - 8 = 28 - 8 = 20$ ✓, matching the point $(4,20)$.

Final answers

• (a) $\dfrac{dy}{dx} = \boxed{2x - \dfrac{16}{x^2}}$
• (b) Stationary point: $\boxed{(2, 12)}$, a minimum point
• (c) Tangent: $\boxed{y = 7x - 8}$