To evaluate $f'(4)$ for the function given in the image:

$f(x) = \frac{(4x^2 - 6)(x^2 - 2)}{x^2 + 4}$

we need to apply differentiation. The function is a quotient, so we'll use the quotient rule. The quotient rule states:

$\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$

where $u(x) = (4x^2 - 6)(x^2 - 2)$ and $v(x) = x^2 + 4$.

Step 1: Differentiate $u(x)$ and $v(x)$

Let's find $u'(x)$ first. To differentiate $u(x) = (4x^2 - 6)(x^2 - 2)$, we'll use the product rule, which is:

$\left(u(x)v(x)\right)' = u'(x)v(x) + u(x)v'(x)$

where:

$u(x) = 4x^2 - 6 \quad \text{and} \quad v(x) = x^2 - 2$

Differentiating each:

$u'(x) = \frac{d}{dx}(4x^2 - 6) = 8x$

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

So,

$u'(x) = 8x(x^2 - 2) + 2x(4x^2 - 6) = 8x(x^2 - 2) + 2x(4x^2 - 6)$

Simplify:

$u'(x) = 8x^3 - 16x + 8x^3 - 12x = 16x^3 - 28x$

Next, differentiate $v(x)$:

$v'(x) = \frac{d}{dx}(x^2 + 4) = 2x$

Step 2: Apply the quotient rule

Now, substitute the derivatives and the original functions into the quotient rule:

$f'(x) = \frac{(16x^3 - 28x)(x^2 + 4) - (4x^2 - 6)(x^2 - 2) \cdot 2x}{(x^2 + 4)^2}$

Step 3: Evaluate at $x = 4$

Substitute $x = 4$ into the expression for $f'(x)$:

1. Calculate $u(4)$, $u'(4)$, $v(4)$, and $v'(4)$.
2. Substitute these values into the derivative formula and simplify.

First, we calculate each component:

• $u(4) = (4(4)^2 - 6)((4)^2 - 2) = (64 - 6)(16 - 2) = 58 \times 14 = 812$
• $u'(4) = 16(4)^3 - 28(4) = 16(64) - 112 = 1024 - 112 = 912$
• $v(4) = 4^2 + 4 = 16 + 4 = 20$
• $v'(4) = 2(4) = 8$

Now:

$f'(4) = \frac{(912)(20) - (812)(8)}{(20)^2} = \frac{18240 - 6496}{400} = \frac{11744}{400} = 29.36$

So, $f'(4) \approx 29.36$.

Would you like more details on any of the steps? Here are some related questions:

1. How is the quotient rule applied in differentiation?
2. What is the product rule in differentiation?
3. Why do we need to simplify the expression before evaluating at a specific point?
4. Can we confirm the result using a different method, like numerical approximation?
5. How does the chain rule relate to more complex derivatives like this one?

Tip: Always simplify the function as much as possible before differentiating to make the differentiation process easier.