Review question

# Can we show $x>\ln(1+x^2)$ for all $x>0$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7431

## Solution

Given that $y=x-\ln(1+x^2),$ show that $\frac{dy}{dx}\geq 0\ \text{for all values of x.}$

We can compute $\dfrac{dy}{dx}$ using the chain rule: \begin{align*} \frac{dy}{dx} &= 1-\frac{1}{1+x^2}\times\frac{d}{dx}(1+x^2)\\ &= 1-\frac{2x}{1+x^2}\\ &= \frac{1+x^2-2x}{1+x^2}\\ &= \frac{(x-1)^2}{1+x^2}. \end{align*}

We know that $(x-1)^2 \geq 0$ and $1+x^2\geq 1$ for all values of $x$, so $\dfrac{dy}{dx} \geq 0$ for all values of $x$.

Deduce that, for all $x>0$, $x>\ln(1+x^2).$

We know that $\dfrac{dy}{dx}\geq 0$ for all values of $x$, so the function $x-\ln(1+x^2)$ is an increasing function.

Substituting we find that when $x = 0$, $y=0$ and $\dfrac{dy}{dx} = 1$. This means that for any value of $x>0$, $y$ must be greater than zero. So \begin{align*} x &- \ln(1+x^2) > 0\\ \implies\quad x &> \ln(1+x^2) \end{align*}

for all $x>0$ as required.