Given that $y = \frac{16x}{1+x^2}$, find the range of values of $x$ for which $y$ is increasing. Another curve has equation $y = \frac{16x}{1+x^2} + d$, where $d$ is a constant. Explain why the range of values of $x$ for which $y$ is increasing for this curve is the same as the answer in (i).

1. Find the derivative of $y$ with respect to $x$ using the quotient rule

Let $u = 16x$ and $v = 1+x^2$.
Then $\frac{du}{dx} = 16$ and $\frac{dv}{dx} = 2x$.
$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{(1+x^2)(16) - (16x)(2x)}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{16 + 16x^2 - 32x^2}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{16 - 16x^2}{(1+x^2)^2}$

2. Set up the inequality for an increasing function

For $y$ to be increasing, $\frac{dy}{dx} > 0$.
$\frac{16(1 - x^2)}{(1+x^2)^2} > 0$

3. Solve the inequality to find the range of $x$

Since $(1+x^2)^2 > 0$ for all real values of $x$, the denominator is always positive.
Therefore, the numerator must be positive:
$16(1 - x^2) > 0$
$1 - x^2 > 0$
$(1 - x)(1 + x) > 0$
$-1 < x < 1$

1. Explain that the derivative of the new curve is identical to the original curve

For $y = \frac{16x}{1+x^2} + d$, differentiating with respect to $x$ gives $\frac{dy}{dx} = \frac{d}{dx}\left(\frac{16x}{1+x^2}\right) + 0$, since the derivative of a constant $d$ is $0$.

2. Conclude that the condition for an increasing function remains unchanged

Since $\frac{dy}{dx}$ is identical for both curves, the inequality $\frac{dy}{dx} > 0$ yields the exact same range of values for $x$.