Differentiation: Question 2

Worked solution

Part (a): Differentiating with the chain rule

Let $u = 2x - 3$, so $y = u^3$.

$\frac{dy}{du} = 3u^2, \qquad \frac{du}{dx} = 2$

By the chain rule,

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = 3u^2 \times 2 = 6u^2 = 6(2x-3)^2$

Independent check, expand and differentiate directly. Using $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a=2x$, $b=3$:

$(2x-3)^3 = 8x^3 - 36x^2 + 54x - 27$

Differentiating term by term:

$\frac{dy}{dx} = 24x^2 - 72x + 54$

Expanding the chain-rule answer to compare: $6(2x-3)^2 = 6(4x^2 - 12x + 9) = 24x^2 - 72x + 54$. The two methods agree exactly, confirming

$\frac{dy}{dx} = 6(2x-3)^2.$

Part (b): Gradient and tangent at $x = 2$

The $y$-coordinate at $x=2$:

$y = (2(2) - 3)^3 = (4-3)^3 = 1^3 = 1$

so the point is $(2, 1)$.

The gradient at $x=2$:

$\frac{dy}{dx}\bigg|_{x=2} = 6(2(2)-3)^2 = 6(1)^2 = 6$

Check using the expanded form from Part (a): $24(2)^2 - 72(2) + 54 = 96 - 144 + 54 = 6$. This matches, confirming the gradient is $6$.

The tangent at $(2,1)$ with gradient $6$:

$y - 1 = 6(x - 2) \implies y = 6x - 12 + 1 = 6x - 11$

Part (c): Normal at $x = 2$

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of $6$:

$m_{\text{normal}} = -\frac{1}{6}$

Using the point $(2,1)$:

$y - 1 = -\frac{1}{6}(x - 2)$

Multiply both sides by $6$ to clear the fraction:

$6y - 6 = -(x - 2) = -x + 2$

$x + 6y = 8$

Check: substituting the point $(2,1)$ back in, $2 + 6(1) = 8$ ✓. Also, the product of the tangent and normal gradients is $6 \times \left(-\tfrac16\right) = -1$, confirming they are indeed perpendicular.

Final answers

• (a) $\dfrac{dy}{dx} = \boxed{6(2x-3)^2}$
• (b) Gradient $= 6$ at $(2,1)$; tangent: $\boxed{y = 6x - 11}$
• (c) Normal: $\boxed{x + 6y = 8}$