This question is to emphasize how our differentiation rules can be used even when it doesn't look like you know enough about a function.  (This is a handy trick to use when your formula is very long to write out.)

Set  h(x)=1+f(x)g(x)
  , where  f
   and  g
   are differentiable functions that satisfy

f(−3)=6
 ,

f′(−3)=4
 ,

g(−3)=5
 , and

g′(−3)=2
 .

Write down a formula for  h′(x)
  on paper.  Using this and the information above, find the value of  h′(−3)
 .
Answer:   h′(−3)=
   

To find \( h'(x) \) where \( h(x) = 1 + f(x)g(x) \), we can use the product rule of differentiation. The product rule states that if \( h(x) = f(x)g(x) \), then \( h'(x) = f'(x)g(x) + f(x)g'(x) \).

Given that \( h(x) = 1 + f(x)g(x) \), the constant 1 has a derivative of 0. Therefore, we only need to differentiate \( f(x)g(x) \).

So, \( h'(x) = \frac{d}{dx}[1 + f(x)g(x)] = \frac{d}{dx}[f(x)g(x)] \).

Applying the product rule, we get:
\[ h'(x) = f'(x)g(x) + f(x)g'(x) \]

Now, we need to evaluate \( h'(x) \) at \( x = -3 \).

Given:
\[ f(-3) = 6 \]
\[ f'(-3) = 4 \]
\[ g(-3) = 5 \]
\[ g'(-3) = 2 \]

Substitute these values into the derivative formula:
\[ h'(-3) = f'(-3)g(-3) + f(-3)g'(-3) \]
\[ h'(-3) = (4)(5) + (6)(2) \]
\[ h'(-3) = 20 + 12 \]
\[ h'(-3) = 32 \]

Thus, the value of \( h'(-3) \) is \( 32 \).

**Answer:** \( h'(-3) = 32 \)

Would you like more details or have any questions about this solution?

Here are 5 related questions for further practice:
1. If \( k(x) = 2 + f(x)g(x) \), how would you find \( k'(-3) \) using the given values of \( f \) and \( g \) and their derivatives?
2. What is the derivative of \( h(x) = f(x)g(x) + x \) at \( x = -3 \)?
3. How would the product rule apply if \( h(x) = f(x)g(x) \cdot h(x) \)?
4. How can you verify the result of \( h'(-3) = 32 \) by using numerical differentiation techniques?
5. How would you apply the chain rule to differentiate \( h(x) = f(g(x)) \)?

**Tip:** When differentiating a product of two functions, remember to apply the product rule carefully and substitute the known values at the specific point to find the derivative at that point.

Set h(x) = 1 + f(x)g(x), where f and g are differentiable functions that satisfy f(−3)=6, f′(−3)=4, g(−3)=5, and g′(−3)=2. Write down a formula for h′(x) on paper. Using this and the information above, find the value of h′(−3).

Learn how to differentiate h(x) = 1 + f(x)g(x) using the product rule of differentiation. This problem involves using given values of f and g at x = -3 to calculate h'(x) and is ideal for students in Grades 11-12. The solution applies the product rule formula and evaluates the derivative at a specific point.

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