Problem 39 - Thomas Calculus

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 \(x=0\). a. \(\frac{d}{d x}(u v)\) b. \(\frac{d}{d x}\left(\frac{u}{v}\right)\) c. \(\frac{d}{d x}\left(\frac{v}{u}\right)\) 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: Use the Product Rule for Part a

To find \( \frac{d}{dx}(uv) \) at \( x=0 \), we apply the product rule for derivatives: \( (uv)' = u'v + uv' \). Substituting the given values:\[(uv)'(0) = u'(0)v(0) + u(0)v'(0) = (-3)(-1) + (5)(2)\]Calculate the terms: \( 3 + 10 = 13 \). So, \( \frac{d}{dx}(uv) = 13 \) at \( x=0 \).

Step 2: Use the Quotient Rule for Part b

To find \( \frac{d}{dx}\left(\frac{u}{v}\right) \) at \( x=0 \), we use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). Applying the given values:\[\left(\frac{u}{v}\right)'(0) = \frac{(-3)(-1) - (5)(2)}{(-1)^2}\]Calculate the numerator: \( 3 - 10 = -7 \), and the denominator: \( 1 \). So, \( \frac{d}{dx}\left(\frac{u}{v}\right) = -7 \) at \( x=0 \).

Step 3: Use the Quotient Rule for Part c

To find \( \frac{d}{dx}\left(\frac{v}{u}\right) \) at \( x=0 \), apply the quotient rule: \( \left(\frac{v}{u}\right)' = \frac{v'u - vu'}{u^2} \). Substituting:\[\left(\frac{v}{u}\right)'(0) = \frac{(2)(5) - (-1)(-3)}{(5)^2}\]This simplifies to \( \frac{10 - 3}{25} = \frac{7}{25} \). Therefore, \( \frac{d}{dx}\left(\frac{v}{u}\right) = \frac{7}{25} \) at \( x=0 \).

Step 4: Differentiate the Linear Combination for Part d

To find \( \frac{d}{dx}(7v - 2u) \) at \( x=0 \), use linearity of differentiation: \( (c_1v + c_2u)' = c_1v' + c_2u' \). Here, \( 7v' - 2u' \):\[7v'(0) - 2u'(0) = 7(2) - 2(-3)\]Calculate: \( 14 + 6 = 20 \). Thus, \( \frac{d}{dx}(7v - 2u) = 20 \) at \( x=0 \).

Key Concepts

Product Rule

When differentiating a product of two functions, such as \( u(x) \) and \( v(x) \), we use the product rule. It helps us find the derivative of two multiplied differentiable functions. If both functions depend on \( x \), the product rule states that the derivative of their product is:\[(uv)' = u'v + uv'\]This means, to get the derivative of the product \( uv \), we take each function and multiply by the derivative of the other, then add these results together.
In our example, substituting the given values:

• \( u'(0) = -3 \), \( v(0) = -1 \)
• \( u(0) = 5 \), \( v'(0) = 2 \)

We compute:\[ (uv)'(0) = (-3)(-1) + (5)(2) = 3 + 10 = 13 \]The result of 13 indicates the rate of change of \( uv \) at \( x = 0 \). This explains how the product rule efficiently provides us with the solution for differentiating products of functions.

Quotient Rule

Calculating the derivative of a quotient involves using the quotient rule, which is essential when dealing with divisions of two differentiable functions. If you have two functions \( u(x) \) and \( v(x) \), their quotient's derivative is given by:\[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\]This formula requires subtracting \( uv' \) (product of the denominator's derivative and the numerator) from \( u'v \) (product of the numerator's derivative and the denominator), and dividing by the square of the denominator.
Applying it to our problem, for \( \frac{u}{v} \) at \( x=0 \):

• \( u'(0) = -3 \), \( v(0) = -1 \)
• \( u(0) = 5 \), \( v'(0) = 2 \)

We find:\[\left(\frac{u}{v}\right)'(0) = \frac{(-3)(-1) - (5)(2)}{(-1)^2} = \frac{3 - 10}{1} = -7\]Similarly, for \( \frac{v}{u} \), we calculate:\[\left(\frac{v}{u}\right)'(0) = \frac{(2)(5) - (-1)(-3)}{(5)^2} = \frac{10 - 3}{25} = \frac{7}{25}\]These calculations underline the versatility of the quotient rule in finding derivatives of functional fractions.

Differentiable Functions

A function is said to be differentiable at a point when it possesses a derivative at that point. This implies the function is smooth and has no breaks, corners, or cusps at that location. Differentiability is a crucial concept because it guarantees that the function has a well-defined rate of change at every point where it is differentiable.
In the given problem, both functions \( u(x) \) and \( v(x) \) are differentiable at \( x=0 \), meaning:

• Their derivatives \( u'(x) \) and \( v'(x) \) exist.
• They possess continuous rates of change at \( x=0 \).

Having these differentiable functions allows us to apply rules like the product and quotient rule effectively. In general, if a function is not differentiable at a certain point, using these standard differentiation techniques would not be possible.

Linearity of Differentiation

Linearity of differentiation refers to the property that allows breaking down the derivative of a linear combination of functions into simpler parts. This property implies that the derivative of a sum (or difference) of functions is simply the sum (or difference) of their derivatives. Mathematically, if \( c_1 \) and \( c_2 \) are constants and \( u(x) \) and \( v(x) \) are functions, then:\[(c_1u + c_2v)' = c_1u' + c_2v'\]In practice, this means we can treat each term independently when differentiating a linear combination.
For our specific problem involving \( 7v - 2u \), substituting the given derivatives:

• \( v'(0) = 2 \), \( u'(0) = -3 \)
• Calculate as: \( 7v'(0) - 2u'(0) = 7(2) - 2(-3) = 14 + 6 = 20 \)

This calculation showcases how using linearity simplifies finding the derivative of combinations of functions, providing a clear and direct solution.