<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show, algebraically, that the derivative of the function $\ln \left( 1+x \right)-\frac{2x}{x+2}$ is never negative.</span></div><p style="text-align: right;">[4]</p></li><li><div class="list-item-content"><span class="list-item-text">Hence show that $\ln \left( 1+x \right)\ge \frac{2x}{x+2}$ when $x\ge 0$.</span></div><p style="text-align: right;">[2]</p></li></ol>

Show, algebraically, that the derivative of the function $\ln \left( 1+x \right)-\frac{2x}{x+2}$ is never negative.

Hence show that $\ln \left( 1+x \right)\ge \frac{2x}{x+2}$ when $x\ge 0$.

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Derivatives of Natural Logarithmic Functions

Techniques of Differentiation

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

IB AA HL Math

Differentiation

9233 P2 Q5 [Modified]

Note: $\dfrac{2x}{x+2} = 2 - \dfrac{4}{x+2}$, so $y = \ln(1+x) + \dfrac{4}{x+2} - 2$

$\dfrac{dy}{dx} = \dfrac{1}{1+x} - \dfrac{4}{(x+2)^2}$

$= \dfrac{(x+2)^2 - 4(x+1)}{(1+x)(x+2)^2}$

$= \dfrac{x^2 + 4x + 4 - 4x - 4}{(x+1)(x+2)^2} = \dfrac{x^2}{(x+1)(x+2)^2}$

$= \dfrac{1}{x+1}\left(\dfrac{x}{x+2}\right)^2$

Since $1 + x > 0$ (for $\ln(1+x)$ to be defined) and $\left(\dfrac{x}{x+2}\right)^2 \geq 0$,

$\therefore \dfrac{dy}{dx} \geq 0$, i.e., the derivative is never negative. (shown)

From (i), $y = \ln(1+x) - \dfrac{2x}{x+2}$ is an increasing function.

Since $y$ is increasing and $y(0) = 0$, for all $x \geq 0$: $y \geq 0$.

$\therefore \ln(1+x) - \dfrac{2x}{x+2} \geq 0$, i.e., $\ln(1+x) \geq \dfrac{2x}{x+2}$ when $x \geq 0$. (shown)

9233 P2 Q5 [Modified]

9233 P2 Q5 [Modified]

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