Problem 53 - University Calculus: Early Transcendentals

Question

Suppose \(u\) and \(v\) are functions of \(x\) that are differentiable at \(x=0\) and that $$u(0)=5, \quad u^{\prime}(0)=-3, \quad v(0)=-1, \quad v^{\prime}(0)=2$$ Find the values of the following derivatives at \(x=0\) $$\text { a. } \frac{d}{d x}(u v)$$ $$\text { b. } \frac{d}{d x}\left(\frac{u}{v}\right)$$ $$\text { c. } \frac{d}{d x}\left(\frac{v}{u}\right)$$ $$\text { d. } \frac{d}{d x}(7 v-2 u)$$

Step-by-Step Solution

Answer

a. 13, b. -7, c. \(\frac{7}{25}\), d. 20

Step 1: Differentiate Product (Part a)

We need to find \( \frac{d}{dx}(uv) \) using the product rule, which states that if \( u \) and \( v \) are differentiable functions, then \( \frac{d}{dx}(uv) = u'v + uv' \). At \( x = 0 \), substitute the given values: \[ u'(0) = -3, \ u(0) = 5, \ v(0) = -1, \ v'(0) = 2 \] So, \[ \frac{d}{dx}(uv)\bigg|_{x=0} = (-3)(-1) + (5)(2) = 3 + 10 = 13 \]

Step 2: Differentiate Quotient (Part b)

To find \( \frac{d}{dx}\left(\frac{u}{v}\right) \), use the quotient rule: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] Substitute the given values at \( x = 0 \): \[ \frac{d}{dx}\left(\frac{u}{v}\right)\bigg|_{x=0} = \frac{(-3)(-1) - (5)(2)}{(-1)^2} = \frac{3 - 10}{1} = -7 \]

Step 3: Differentiate Reciprocal (Part c)

To find \( \frac{d}{dx}\left(\frac{v}{u}\right) \), also use the quotient rule: \[ \frac{d}{dx}\left(\frac{v}{u}\right) = \frac{v'u - uv'}{u^2} \] Substitute the given values: \[ \frac{d}{dx}\left(\frac{v}{u}\right)\bigg|_{x=0} = \frac{(2)(5) - (-1)(-3)}{5^2} = \frac{10 - 3}{25} = \frac{7}{25} \]

Step 4: Differentiate Linear Combination (Part d)

We need to find \( \frac{d}{dx}(7v - 2u) \). The derivative of a linear combination is simply the linear combination of the derivatives: \[ \frac{d}{dx}(7v - 2u) = 7v' - 2u' \] Using the given values: \[ \frac{d}{dx}(7v - 2u)\bigg|_{x=0} = 7(2) - 2(-3) = 14 + 6 = 20 \]

Key Concepts

Product Rule

When you're working with the product of two functions, such as \(u(x)v(x)\), the product rule comes in handy for differentiation. The rule states: \[\frac{d}{dx}(uv) = u'v + uv'\] This means you'll take the derivative of the first function, multiply it by the second function, and then add it to the first function multiplied by the derivative of the second function. Here's a breakdown of how it works:

• Find the derivative of function \(u\), which is \(u'\).
• Multiply \(u'\) by the function \(v\).
• Find the derivative of function \(v\), which is \(v'\).
• Multiply \(v'\) by the function \(u\).
• Add these two results together to get the final derivative.

So, the combined effort of differentiating and adding these components gives you the derivative of the product of two functions.

Quotient Rule

The quotient rule is what you use when differentiating a function that's the ratio of two differentiable functions, such as \(\frac{u}{v}\). The quotient rule is given by:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\]This rule requires you to:

• Differentiate the numerator to get \(u'\).
• Differentiate the denominator to get \(v'\).
• Multiply \(u'\) by \(v\) and \(u\) by \(v'\).
• Subtract the second product from the first.
• Divide everything by the square of \(v\).

This approach ensures that you are systematically finding the rate of change of a quotient of two functions.

Reciprocal Function

The concept of a reciprocal function ties closely to the quotient rule. If you have \(\frac{v}{u}\), and you need to differentiate it, you will again employ the quotient rule but apply it to the inverse form: \[\frac{d}{dx}\left(\frac{v}{u}\right) = \frac{v'u - uv'}{u^2}\]Key things to remember:

• You are essentially considering the function as a ratio of two differentiable functions, just reversed.
• Find derivatives just like with the standard quotient rule.
• Pay careful attention to sign changes that can occur in subtraction.
• Divide by the square of the denominator function.

This makes differentiating reciprocal functions systematic and consistent, relying on foundational calculus rules.

Linear Combination

Linear combinations involve sums of functions multiplied by constants. When differentiating a linear combination like \(7v - 2u\), it is quite straightforward:The rule is: \[\frac{d}{dx}(7v - 2u) = 7v' - 2u'\] Here's how it works:

• The derivative of a constant times a function is the constant times the derivative of the function.
• Differentiate each function in the combination independently.
• Multiply the resulting derivatives by their respective constants.
• Combine the results using addition or subtraction exactly as in the original function.

This shows how differentiation respects the linearity of functions, making these calculations simpler to perform.