Differentiation: Question 7

Worked solution

Part (a): Differentiating with the chain rule

Let $u = 2x + 7$, so $y = u^{1/2}$.

$\frac{dy}{du} = \frac12 u^{-1/2}, \qquad \frac{du}{dx} = 2$

By the chain rule,

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac12 u^{-1/2} \times 2 = u^{-1/2} = (2x+7)^{-1/2} = \frac{1}{\sqrt{2x+7}}$

Independent check. Implicit differentiation. Since $y = \sqrt{2x+7}$ with $y > 0$, squaring both sides gives $y^2 = 2x+7$. Differentiating both sides with respect to $x$:

$2y\frac{dy}{dx} = 2 \implies \frac{dy}{dx} = \frac{1}{y} = \frac{1}{\sqrt{2x+7}}$

This matches the chain-rule result exactly, confirming

$\frac{dy}{dx} = \frac{1}{\sqrt{2x+7}}.$

Part (b): Gradient and tangent at $x = 1$

The $y$-coordinate at $x=1$:

$y = \sqrt{2(1) + 7} = \sqrt{9} = 3$

so the point is $(1, 3)$.

The gradient at $x=1$:

$\frac{dy}{dx}\bigg|_{x=1} = \frac{1}{\sqrt{9}} = \frac{1}{3}$

Check using the implicit form: $\dfrac{1}{y} = \dfrac13$ at $y=3$ ✓, confirming the gradient is $\tfrac13$.

The tangent at $(1,3)$ with gradient $\tfrac13$:

$y - 3 = \frac13(x - 1) \implies y = \frac13 x - \frac13 + 3 = \frac13 x + \frac83$

Multiply both sides by $3$ to clear the fraction:

$3y = x + 8$

Part (c): Normal at $x = 1$

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of $\tfrac13$:

$m_{\text{normal}} = -3$

Using the point $(1,3)$:

$y - 3 = -3(x - 1) \implies y = -3x + 3 + 3 = -3x + 6$

Rearranging:

$3x + y = 6$

Check: substituting the point $(1,3)$ back in, $3(1) + 3 = 6$ ✓. Also, the product of the tangent and normal gradients is $\tfrac13 \times (-3) = -1$, confirming they are indeed perpendicular.

Final answers

• (a) $\dfrac{dy}{dx} = \boxed{\dfrac{1}{\sqrt{2x+7}}}$
• (b) Gradient $= \tfrac13$ at $(1,3)$; tangent: $\boxed{3y = x + 8}$
• (c) Normal: $\boxed{3x + y = 6}$