Questions: Let h(x)=f(x) g(x). If f(x)=x^2-1 and g(x)=-x^2-4 x-2, what is h'(−2)?

Transcript text: Let $h(x)=f(x) g(x)$. If $f(x)=x^{2}-1$ and $g(x)=-x^{2}-4 x-2$, what is $h^{\prime}(-2) ?$

Solution

Solution Steps

To find the derivative of the function $ h(x) = f(x)g(x) $ using the product rule, we first need to identify the derivatives of $ f(x) $ and $ g(x) $. The product rule states that the derivative of a product of two functions is given by $ h'(x) = f'(x)g(x) + f(x)g'(x) $. After finding $ h'(x) $, we evaluate it at $ x = -2 $.

Step 1: Define the Functions

We have the functions: \[ f(x) = x^2 - 1 \] \[ g(x) = -x^2 - 4x - 2 \]

Step 2: Compute the Derivatives

The derivatives of the functions are: \[ f'(x) = 2x \] \[ g'(x) = -2x - 4 \]

Step 3: Apply the Product Rule

Using the product rule, we find the derivative of $ h(x) = f(x)g(x) $: \[ h'(x) = f'(x)g(x) + f(x)g'(x) \] Substituting the derivatives and the original functions: \[ h'(x) = (2x)(-x^2 - 4x - 2) + (x^2 - 1)(-2x - 4) \]

Step 4: Evaluate at $ x = -2 $

Now we evaluate $ h'(-2) $: \[ h'(-2) = 2(-2)(-(-2)^2 - 4(-2) - 2) + ((-2)^2 - 1)(-2(-2) - 4) \] Calculating each part: \[ = 2(-2)(-4 + 8 - 2) + (4 - 1)(4 - 4) \] \[ = 2(-2)(2) + 3(0) \] \[ = -8 + 0 = -8 \]