6. (a) Differentiate the function $(2x + 4)^2$ from first principles, with respect to $x$ - Leaving Cert Mathematics - Question 6 - 2016

To differentiate ( y = x \sin \left( \frac{1}{x} \right) ), we use the product rule:

Let ( u = x ) and ( v = \sin \left( \frac{1}{x} \right) ), then:

1. ( u' = 1 )
2. Differentiate ( v ) using chain rule:
   • First, differentiate ( \sin(u) ),\ with ( u = \frac{1}{x} ):
     $v' = \cos \left( \frac{1}{x} \right) \frac{d}{dx} \left( \frac{1}{x} \right) = \cos \left( \frac{1}{x} \right) \left( -\frac{1}{x^2} \right)$

Now use the product rule:

$\frac{dy}{dx} = u'v + uv' = 1 \cdot \sin \left( \frac{1}{x} \right) + x \left( -\frac{1}{x^2} \cos \left( \frac{1}{x} \right) \right)$
This simplifies to:

$\frac{dy}{dx} = \sin \left( \frac{1}{x} \right) - \frac{1}{x} \cos \left( \frac{1}{x} \right)$.

We can find the slope of the tangent by substituting ( x = \frac{4}{\pi} ) into our derivative:

The derivative is: $\frac{dy}{dx} = \sin \left( \frac{1}{x} \right) - \frac{1}{x} \cos \left( \frac{1}{x} \right)$.

Substituting ( x = \frac{4}{\pi} ):
$\frac{dy}{dx} = \sin \left( \frac{\pi}{4} \right) - \frac{\pi}{4} \cos \left( \frac{\pi}{4} \right)$
Using known values:
$\sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \quad \text{and} \quad \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}$,

Thus the slope becomes: $\frac{dy}{dx} = \frac{\sqrt{2}}{2} - \frac{\pi}{4} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \left( 1 - \frac{\pi}{4} \right)$.

Calculating this value gives: $\frac{dy}{dx} \approx 0.15$ (to two decimal places).