Question

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d) \\(y = 4x^4(3x^2 - 1)\\)

\\(u = 4x^4\\) \t \\(v = 3x^2 - 1\\)
\\(u = 16x^3\\) \t \\(v = 6x\\)

\\(y = 6x(4x^4) + 16x^3(3x^2 - 1)\\)

Explanation:

Identify the function and components

The given function is:

$$ y = 4x^4(3x^2 - 1) $$

We can define the two factors as:

$$ u = 4x^4 $$

$$ v = 3x^2 - 1 $$

Find the individual derivatives

Using the power rule for differentiation:

$$ u' = \frac{d}{dx}(4x^4) = 16x^3 $$

$$ v' = \frac{d}{dx}(3x^2 - 1) = 6x $$

Apply the product rule formula

The product rule states:

$$ y' = u v' + v u' $$

Substituting our functions and their derivatives:

$$ y' = (4x^4)(6x) + (3x^2 - 1)(16x^3) $$

This matches the expression written in the image:

$$ y' = 6x(4x^4) + 16x^3(3x^2 - 1) $$

Simplify the derivative expression

Now we expand and combine like terms to find the fully simplified derivative:

$$ y' = 24x^5 + 16x^3(3x^2) - 16x^3 $$

$$ y' = 24x^5 + 48x^5 - 16x^3 $$

$$ y' = 72x^5 - 16x^3 $$

Factor the final simplified result

We can factor out the greatest common factor, which is \(8x^3\):

$$ y' = 8x^3(9x^2 - 2) $$

Answer:

$$ y' = 72x^5 - 16x^3 = 8x^3(9x^2 - 2) $$