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Magazine Chapter 2: Problem 30
Find \(F^{\prime}(\pi)\) given that \(f(\pi)=10, f^{\prime}(\pi)=-1, g(\pi)=-3\) and \(g^{\prime}(\pi)=2\). (a) \(F(x)=6 f(x)-5 g(x)\) (b) \(F(x)=x(f(x)+g(x))\) (c) \(F(x)=2 f(x) g(x)\) (d) \(F(x)=\frac{f(x)}{4+g(x)}\)
Short Answer
Expert verified
(a) -16, (b) 7 + 蟺, (c) 46, (d) -21
Step by step solution
01
Differentiate Part (a) Function
For part (a), we have the function \( F(x) = 6f(x) - 5g(x) \). To find \( F^{\prime}(x) \), use the linearity of differentiation: \( F^{\prime}(x) = 6f^{\prime}(x) - 5g^{\prime}(x) \). Evaluate this at \( x = \pi \): \( F^{\prime}(\pi) = 6f^{\prime}(\pi) - 5g^{\prime}(\pi) = 6(-1) - 5(2) = -6 - 10 = -16 \).
02
Differentiate Part (b) Function
For part (b), the function is \( F(x) = x(f(x) + g(x)) \). Apply the product rule: \( F^{\prime}(x) = 1(f(x) + g(x)) + x(f^{\prime}(x) + g^{\prime}(x)) = f(x) + g(x) + x(f^{\prime}(x) + g^{\prime}(x)) \). Evaluate at \( x=\pi \): \( F^{\prime}(\pi) = f(\pi) + g(\pi) + \pi(f^{\prime}(\pi) + g^{\prime}(\pi)) = 10 - 3 + \pi(-1 + 2) = 7 + \pi \).
03
Differentiate Part (c) Function
For part (c), the function is \( F(x) = 2f(x)g(x) \). Use the product rule: \( F^{\prime}(x) = 2(f^{\prime}(x)g(x) + f(x)g^{\prime}(x)) \). Evaluate this at \( x=\pi \): \( F^{\prime}(\pi) = 2((-1)(-3) + 10(2)) = 2(3 + 20) = 2(23) = 46 \).
04
Differentiate Part (d) Function
For part (d), the function is \( F(x) = \frac{f(x)}{4 + g(x)} \). Use the quotient rule: \( F^{\prime}(x) = \frac{(4 + g(x))f^{\prime}(x) - f(x)g^{\prime}(x)}{(4 + g(x))^2} \). Evaluate this at \( x=\pi \): \( F^{\prime}(\pi) = \frac{(4 - 3)(-1) - 10(2)}{(4 - 3)^2} = \frac{-1 - 20}{1} = -21 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linearity of Differentiation
The linearity of differentiation is a fundamental concept in differential calculus. It refers to the property of derivatives that allows us to differentiate expressions in a component-wise manner. This means we can differentiate each part of an expression separately and then combine the results.
For example, if we have a function like \( F(x) = a f(x) + b g(x) \), where \( a \) and \( b \) are constants, the derivative is
\[ F'(x) = a f'(x) + b g'(x) \].
This is because differentiation follows additive and scalar multiplication rules.
For example, if we have a function like \( F(x) = a f(x) + b g(x) \), where \( a \) and \( b \) are constants, the derivative is
\[ F'(x) = a f'(x) + b g'(x) \].
This is because differentiation follows additive and scalar multiplication rules.
- Additive Rule: The derivative of the sum of two functions is the sum of their derivatives.
- Scalar Rule: The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function.
Product Rule
When you encounter a function that is the product of two other functions, the product rule helps in finding its derivative. The product rule is expressed as:
For functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is given by
\[ (uv)' = u'v + uv' \].
This rule is particularly useful when dealing with expressions where functions are multiplied together, such as \( F(x) = x(f(x) + g(x)) \) or \( F(x) = 2 f(x) g(x) \).
For functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is given by
\[ (uv)' = u'v + uv' \].
This rule is particularly useful when dealing with expressions where functions are multiplied together, such as \( F(x) = x(f(x) + g(x)) \) or \( F(x) = 2 f(x) g(x) \).
- To apply it, differentiate one function while keeping the other constant, then switch the roles, and finally add the results.
- Remember, you must differentiate all functions involved when applying the rule.
Quotient Rule
The quotient rule is essential when dealing with functions that are in the form of a fraction. It allows you to find the derivative of a quotient of two functions. The rule is:
For functions \( u(x) \) and \( v(x) \), the derivative of their quotient \( \frac{u(x)}{v(x)} \) is given by
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \].
This formula helps in deriving expressions like \( F(x) = \frac{f(x)}{4+g(x)} \).
For functions \( u(x) \) and \( v(x) \), the derivative of their quotient \( \frac{u(x)}{v(x)} \) is given by
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \].
This formula helps in deriving expressions like \( F(x) = \frac{f(x)}{4+g(x)} \).
- Start by differentiating the numerator and the denominator separately.
- Apply the formula by substituting into the components as \( u \) and \( v \).
- Finally, the denominator \( v^2 \) remains in the denominator of the derivative, providing a clear difference from the product rule.
Functional Evaluation
Functional evaluation involves inserting a specific value into a function or its derivative. It helps us find the precise rate of change at a specific point. For example, in our exercise, we need to evaluate \( F'( \pi) \).
- This involves plugging the specific value provided, typically \( \pi \) in this case, into the differentiated equation.
- Calculating the result allows us to understand how the function behaves exactly at that point.
- By knowing values like \( f(\pi) \) and \( f'(\pi) \), it becomes easier to calculate results accurately without re-evaluating the entire function.
