Question:

Let $f(x)$ and $g(x)$ be differentiable functions such that $f(2) = 4$, $g(2) = 3$, $f'(2) = 5$, and $g'(2) = -2$. Given that $h(x) = (f \circ g)(x)$, find the value of $h'(2)$.

Answer:

To find $h'(2)$, we will use the chain rule. The chain rule states that if we have a function $h(x) = (f \circ g)(x)$, then its derivative $h'(x)$ can be calculated as the product of the derivative of the outer function $f'(g(x))$ and the derivative of the inner function $g'(x)$. In equation form:

Now let's find $h'(2)$ step by step.

Step 1: Evaluate $f'(g(2))$ According to the chain rule, we need to evaluate $f'(g(x))$ and substitute $x = 2$. Since $g(2) = 3$, we have $f'(g(2)) = f'(3)$.

Step 2: Evaluate $g'(2)$ We are given that $g'(2) = -2$.

Step 3: Multiply the results from Step 1 and Step 2 Multiply $f'(g(2))$ and $g'(2)$. This will give us $f'(3) \cdot (-2)$.

Step 4: Calculate $f'(3)$ We are given that $f'(2) = 5$. To calculate $f'(3)$, we need to use the fact that $f'(x)$ represents the rate of change of $f(x)$ with respect to $x$. Therefore, if $f'(2) = 5$, it means that for every unit increase in $x$ (in this case from 2 to 3), $f(x)$ increases by 5 units. So, $f'(3) = 5$.

Step 5: Substitute values into the equation Substituting $f'(3) = 5$ and $g'(2) = -2$ into the equation from Step 3, we have:

Therefore, $h'(2) = -10$. The value of the derivative of $h(x)$ at $x = 2$ is $-10$.

Hence, $h'(2) = \boxed{-10}$.