(a) Y = (x – 2)(x + 2)(x – 3)(x + 4)(b) * Y = 8 Log x

Let’s differentiate the given functions with respect to ( x ).

(a) ( Y = (x – 2)(x + 2)(x – 3)(x + 4) )

We need to use the product rule multiple times. Let:
[ Y = f(x) \cdot g(x) ]
where
[ f(x) = (x – 2)(x + 2) ]
[ g(x) = (x – 3)(x + 4) ]

First, differentiate ( f(x) ) and ( g(x) ) individually:
[ f(x) = (x – 2)(x + 2) = x^2 – 4 ]
[ f'(x) = 2x ]

[ g(x) = (x – 3)(x + 4) = x^2 + x – 12 ]
[ g'(x) = 2x + 1 ]

Now apply the product rule to differentiate ( Y ):
[ Y = f(x) \cdot g(x) ]
[ \frac{dY}{dx} = f'(x) \cdot g(x) + f(x) \cdot g'(x) ]

Substitute the expressions for ( f(x), f'(x), g(x), ) and ( g'(x) ):
[ \frac{dY}{dx} = (2x) \cdot (x^2 + x – 12) + (x^2 – 4) \cdot (2x + 1) ]

Simplify each term:
[ \frac{dY}{dx} = 2x(x^2 + x – 12) + (x^2 – 4)(2x + 1) ]
[ = 2x^3 + 2x^2 – 24x + 2x^3 – 8x + x^2 – 4 ]
[ = 4x^3 + 3x^2 – 32x – 4 ]

So, the derivative of ( Y ) with respect to ( x ) is:
[ \frac{dY}{dx} = 4x^3 + 3x^2 – 32x – 4 ]

(b) ( Y = 8 \log x )

To differentiate ( Y = 8 \log x ) with respect to ( x ), use the fact that the derivative of ( \log x ) (where the base is the natural logarithm, ( \log_e x )) is ( \frac{1}{x} ):

[ \frac{dY}{dx} = 8 \cdot \frac{d}{dx} (\log x) ]
[ \frac{dY}{dx} = 8 \cdot \frac{1}{x} ]
[ \frac{dY}{dx} = \frac{8}{x} ]

So, the derivative of ( Y ) with respect to ( x ) is:
[ \frac{dY}{dx} = \frac{8}{x} ]