Question

Given that \(f(1)=2, f^{\prime}(1)=-1, g(1)=0 \quad\) and \(g^{\prime}(1)=1\), find \(F^{\prime}(1)\) where \(F(x)=f(x) \cos g(x)\).

Short answer

The value of \( F'(1) \) is \(-1\).

Step-by-step solution

Understanding the Chain Rule for Derivatives

To find the derivative of the function \( F(x) = f(x) \cdot \cos(g(x)) \), we need to apply the product rule along with the chain rule.\ The product rule states that if we have a function \( u(x) \cdot v(x) \), then the derivative is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Here, \( u(x) = f(x) \) and \( v(x) = \cos(g(x)) \).

Applying the Product Rule

From the product rule, \( F'(x) = f'(x) \cdot \cos(g(x)) + f(x) \cdot (-\sin(g(x)) \cdot g'(x)) \). \ This comes from differentiating \( v(x) = \cos(g(x)) \), which requires the chain rule: \( v'(x) = -\sin(g(x)) \cdot g'(x) \).

Substitute Known Values into the Derivative

Now substitute the given values: \( f(1) = 2, f'(1) = -1, g(1) = 0, g'(1) = 1 \) into the derivative obtained.\ \[ F'(1) = (f'(1) \cdot \cos(g(1)) + f(1) \cdot (-\sin(g(1)) \cdot g'(1))) \].

Simplify the Expression

With \( g(1) = 0 \), we know that \( \cos(0) = 1 \) and \( \sin(0) = 0 \). Substitute these into the expression: \[ F'(1) = (-1) \cdot 1 + 2 \cdot (0 \cdot 1) \].

Calculate the Result

Substitute the calculated values: \[ F'(1) = -1 + 0 \]. Thus, the final value is \[ F'(1) = -1 \].

Key concepts

Product Rule

The product rule is a fundamental tool in calculus used for finding the derivative of products of two or more functions. Imagine you have two functions, say, \( u(x) \) and \( v(x) \). The product of these functions is \( u(x) \cdot v(x) \), and their derivative isn't as simple as differentiating each one separately and then multiplying the results. Instead, we need a special rule, the product rule.

The product rule formula is:

• \( (u \cdot v)' = u' \cdot v + u \cdot v' \)
• This tells us to take the derivative of the first function, multiply it by the second function, and add the result to the first function multiplied by the derivative of the second function.

This rule is especially handy to correctly differentiate expressions where functions are multiplied together. It helps ensure that we account for changes in both functions at the same time. Seeing it in action with our function \( f(x) \cdot \cos(g(x)) \) shows just how useful it can be in combination with other rules like the chain rule.

Chain Rule

The chain rule is another essential calculus tool, used for differentiating compositions of functions. It comes into play when one function is nested inside another, creating a composite function. In simple terms, if you have a function \( h(x) = f(g(x)) \), then the chain rule helps us find the derivative of this composite function.

The chain rule formula is:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)
• This essentially means you first differentiate the outer function \( f \) with respect to the inner function \( g \), then multiply by the derivative of the inner function \( g \).

In our example, the chain rule was used when differentiating \( \cos(g(x)) \). The cosine function is the outer function, and \( g(x) \) is the inner one. Applying the chain rule here gave us \( -\sin(g(x)) \cdot g'(x) \), which is critical to completing the derivative process accurately.

Function Differentiation

Differentiation is the process of finding the derivative, which tells us how a function changes at any given point. It's central to calculus and allows us to solve many real-world problems related to rates of change. A derivative gives the slope of a function at a particular point, helping us analyze trends and relationships between variables.

When differentiating a function, various rules can be applied, depending on the function's structure. The product rule and chain rule are among these, as they've demonstrated their necessity in our example function \( F(x) = f(x) \cdot \cos(g(x)) \).

We used both rules because our function involved a product and a composition of functions. By correctly applying these rules, we ensured that each part of the initial function \( F(x) \) was differentiated accurately, giving us the derivative \( F'(1) \) which was calculated to be \(-1\). Mastery of function differentiation involves recognizing which rules to apply and breaking down the function into manageable parts.