Consider the following for the two (02) items that follow: Let x =...

Given $x = \sec\theta - \cos\theta$ and $y = \sec^4\theta - \cos^4\theta$.

First, we simplify $x^2 + 4 and y^2 + 4$:
$x^2 + 4 = (\sec\theta - \cos\theta)^2 + 4 = \sec^2\theta + \cos^2\theta - 2 + 4 = (\sec\theta + \cos\theta)^2$
$y^2 + 4 = (\sec^4\theta - \cos^4\theta)^2 + 4 = \sec^8\theta + \cos^8\theta - 2 + 4 = (\sec^4\theta + \cos^4\theta)^2$

Differentiating $x and y$ with respect to $\theta$:
$\dfrac{dx}{d\theta} = \sec\theta\tan\theta + \sin\theta = \tan\theta(\sec\theta + \cos\theta)$
$\dfrac{dy}{d\theta} = 4\sec^3\theta(\sec\theta\tan\theta) - 4\cos^3\theta(-\sin\theta) = 4\tan\theta(\sec^4\theta + \cos^4\theta)$

Now, finding $\dfrac{dy}{dx}$:
$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{4\tan\theta(\sec^4\theta + \cos^4\theta)}{\tan\theta(\sec\theta + \cos\theta)} = \dfrac{4(\sec^4\theta + \cos^4\theta)}{\sec\theta + \cos\theta}$

Squaring both sides:
\left(\dfrac{dy}{dx}\right)^2 = \dfrac{16(\sec^4\theta + \cos^4\theta)^2}{(\sec\theta + \cos\theta)^2} = \dfrac{16(y^2+4)}{x^2+4}$<br><br>Rearranging the equation:<br>$(x^2+4)\left(\dfrac{dy}{dx}\right)^2 = 16(y^2+4)$<br><br>Differentiating both sides with respect to$x$:<br>$2x\left(\dfrac{dy}{dx}\right)^2 + (x^2+4) \cdot 2\dfrac{dy}{dx} \cdot \dfrac{d^2y}{dx^2} = 16 \cdot 2y\dfrac{dy}{dx}$<br><br>Dividing the entire equation by$2\dfrac{dy}{dx}$:<br>$x\dfrac{dy}{dx} + (x^2+4)\dfrac{d^2y}{dx^2} = 16y$<br>$\Rightarrow (x^2+4)\dfrac{d^2y}{dx^2} - 16y = -x\dfrac{dy}{dx}<br><br>Substituting this into the given expression:<br>\left(\dfrac{x^2+4}{y^2+4}\right)\dfrac{dy}{dx}\left[(x^2+4)\dfrac{d^2y}{dx^2} - 16y\right] = \left(\dfrac{x^2+4}{y^2+4}\right)\dfrac{dy}{dx}\left(- x\dfrac{dy}{dx}\right)<br><br>This simplifies to:<br>-x\left(\dfrac{x^2+4}{y^2+4}\right)\left(\dfrac{dy}{dx}\right)^2$<br><br>Substituting \left(\dfrac{dy}{dx}\right)^2 = \dfrac{16(y^2+4)}{x^2+4} into$ the expression:
-x\left(\dfrac{x^2+4}{y^2+4}\right)\left(\dfrac{16(y^2+4)}{x^2+4}\right) = -16x$<br><br>Answer:$-16x