Given that f(1) = 2, f'(1) = -1, g(1) = 0, and g'(1) = 1. If F(x) = f(x) cos cos (g (x)), then the value of F'(1) is?"

To find the value of $F'(1)$ for the function $F(x) = f(x) \cos(\cos(g(x)))$, we will use the product rule and the chain rule of differentiation.

1. Break Down the Problem

We need to differentiate $F(x)$ and evaluate it at $x = 1$. The function can be expressed as a product: $F(x) = u(x) \cdot v(x)$ where:

• $u(x) = f(x)$
• $v(x) = \cos(\cos(g(x)))$

2. Relevant Concepts

Using the product rule for differentiation: $F'(x) = u'(x)v(x) + u(x)v'(x)$ We will need to calculate $u'(x) = f'(x)$ and $v'(x)$. For $v(x)$, we apply the chain rule: $v(x) = \cos(\cos(g(x))) \quad \Rightarrow \quad v'(x) = -\sin(\cos(g(x))) \cdot \sin(g(x)) \cdot g'(x)$

3. Analysis and Detail

Now, we compute $u'(1), v(1),$ and $v'(1)$:

• $u'(1) = f'(1) = -1$
• $u(1) = f(1) = 2$
• $g(1) = 0$
• $g'(1) = 1$

Now calculate $v(1)$: $v(1) = \cos(\cos(g(1))) = \cos(\cos(0)) = \cos(1)$

Now, calculate $v'(1)$: $v'(1) = -\sin(\cos(g(1))) \cdot \sin(g(1)) \cdot g'(1) = -\sin(\cos(0)) \cdot \sin(0) \cdot 1 = -\sin(1) \cdot 0 = 0$

4. Verify and Summarize

Now substituting everything back into the equation for $F'(1)$: $F'(1) = u'(1)v(1) + u(1)v'(1) = (-1)(\cos(1)) + (2)(0) = -\cos(1)$

Final Answer

The value of $F'(1)$ is: $F'(1) = -\cos(1)$