# Differentation (Normal line/ Approximation)

\CIE\0606\2020\w\paper 12 \no 7
A curve has equation $y=\dfrac{\ln \left(3 x^{2}-5\right)}{2 x+1}$ for $3 x^{2}>5$.
(a) Find the equation of the normal to the curve at the point where $x=\sqrt{2}$.
(b) Find the approximate change in $y$ as $x$ increases from $\sqrt{2}$ to $\sqrt{2}+h$, where $h$ is   small.

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$\def\dbydxq#1#2{\dfrac{(#2)\dfrac{d}{dx}(#1)-#1\dfrac{d}{dx}(#2)}{(#2)^2}}$
(a) When $x=\sqrt 2,y=\dfrac{\ln (3\times 2-5)}{2\sqrt 2+1}=0.$
$\begin{eqnarray*}y&=&\dfrac{\ln (3x^2-5)}{2x+1}\\\dfrac{dy}{dx}&=&\dbydxq{\ln (3x^2-5)}{2x+1}\\&=&\dfrac{(2x+1)\dfrac{6x}{3x^2-5}-\ln(3x^2-5)(2)}{(2x+1)^2}\end{eqnarray*}$
When $x=\sqrt 2,$
$\begin{eqnarray*}\dfrac{dy}{dx}&=&\dfrac{(2\sqrt 2+1)\dfrac{6(\sqrt 2)}{3(\sqrt{2})^2-5}-\ln(3(\sqrt 2)^2-5)(2)}{(2(\sqrt 2)+1)^2}\\&=&\dfrac{6\sqrt 2}{2\sqrt 2+1}=\dfrac{24-6\sqrt 2}{7}\end{eqnarray*}$
$\therefore$ the gradient of normal line is $-\dfrac{(2\sqrt 2+1)}{6\sqrt 2}.$
Normal line equation: $y=-\dfrac{(2\sqrt 2+1)}{6\sqrt 2}(x-\sqrt 2).$
(b) $\Delta y=\dfrac{dy}{dx}\times \Delta x=\dfrac{24-6\sqrt 2}{7}h.$
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