Differentiate the equation using quotient rule and chain rule.

$u = \ln (3 x^{2} - 5)$ $v = 2 x + 1$

$\frac{d u}{d x} = (\frac{1}{3 x^{2} - 5}) \times 6 x$ $\frac{d v}{d x} = 2$

correct $\frac{d u}{d x}$ [1]

Use quotient rule, $\frac{d y}{d x} = \frac{(v) (\frac{d u}{d x}) - (u) (\frac{d v}{d x})}{v^{2}}$ .

$\frac{d y}{d x} = \frac{((2 x + 1) (\frac{6 x}{3 x^{2} - 5})) - (2) (\ln (3 x^{2} - 5))}{{(2 x + 1)}^{2}}$

quotient rule [1]
all terms correct except $\frac{6 x}{3 x^{2} - 5}$ [1]

When $x = \sqrt{2}$ ,

$y = \frac{\ln (1)}{2 \sqrt{2} + 1}$

$\ln (1) = 0$ , therefore when $x = \sqrt{2}$ ,

$y = 0$

[1]

Substitute $x = \sqrt{2}$ into $\frac{d y}{d x}$ to find the gradient of the curve at this point.

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{(2 \sqrt{2} + 1) (\frac{6 \sqrt{2}}{3 {(\sqrt{2})}^{2} - 5}) - 2 \ln (3 {(\sqrt{2})}^{2} - 5)}{{(2 \sqrt{2} + 1)}^{2}} \\ = & \frac{(\frac{(2 \sqrt{2} + 1) (6 \sqrt{2})}{1}) - 2 \ln (1)}{{(2 \sqrt{2} + 1)}^{2}} \\ = & \frac{(2 \sqrt{2} + 1) (6 \sqrt{2})}{{(2 \sqrt{2} + 1)}^{2}} \\ = & \frac{6 \sqrt{2}}{2 \sqrt{2} + 1} \end{array}$

Take the negative reciprocal to find the gradient of the normal.

$gradient (normal) = - \frac{2 \sqrt{2} + 1}{6 \sqrt{2}}$

Substitute this gradient and the point $(\sqrt{2}, 0)$ into the equation of line, $y - y_{1} = m (x - x_{1})$ .

$\begin{array}{rcl} y - 0 & = & - \frac{2 \sqrt{2} + 1}{6 \sqrt{2}} (x - \sqrt{2}) \end{array}$

[1]

$\begin{array}{rcl} y & = & - \frac{(2 \sqrt{2} + 1) (x - \sqrt{2})}{6 \sqrt{2}} \\ y & = & - 0.451 184 . . . x + 0.638 071 . . . \end{array}$

$y = - 0.451 x + 0.638 (3 s . f .)$ [1]

Differentiation (Cambridge O Level Additional Maths)

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