Problem 54 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

a. -2, b. \(\frac{2}{25}\), c. -\(\frac{1}{2}\), d. -7.

Step 1: Use the Product Rule for Part (a)

To find \( \frac{d}{dx}(uv) \), which is the derivative of a product, we apply the Product Rule: \( \frac{d}{dx}(uv) = u'v + uv' \). At \( x = 1 \), substitute \( u(1) = 2 \), \( u'(1) = 0 \), \( v(1) = 5 \), and \( v'(1) = -1 \) into the equation. Thus, we have:\[ u'(1)v(1) + u(1)v'(1) = 0 \cdot 5 + 2 \cdot (-1) = -2 \]. The value is \( -2 \).

Step 2: Apply the Quotient Rule for Part (b)

For \( \frac{d}{dx}\left(\frac{u}{v}\right) \), use the Quotient Rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \). Substituting \( u(1) = 2 \), \( u'(1) = 0 \), \( v(1) = 5 \), and \( v'(1) = -1 \), we get:\[ \frac{0 \cdot 5 - 2 \cdot (-1)}{5^2} = \frac{2}{25} \]. The value is \( \frac{2}{25} \).

Step 3: Use the Quotient Rule for Part (c)

Similarly, for \( \frac{d}{dx}\left(\frac{v}{u}\right) \), apply the Quotient Rule: \( \frac{d}{dx}\left(\frac{v}{u}\right) = \frac{v'u - vu'}{u^2} \). Plug in \( u(1) = 2 \), \( u'(1) = 0 \), \( v(1) = 5 \), and \( v'(1) = -1 \):\[ \frac{-1 \cdot 2 - 5 \cdot 0}{2^2} = \frac{-2}{4} = -\frac{1}{2} \]. The value is \( -\frac{1}{2} \).

Step 4: Differentiate a Linear Combination for Part (d)

For \( \frac{d}{dx}(7v - 2u) \), apply the sum rule: \( 7v' - 2u' \). Using \( u'(1) = 0 \) and \( v'(1) = -1 \), substitute into the equation:\[ 7(-1) - 2(0) = -7 \]. The value is \( -7 \).

Key Concepts

Product Rule

Whenever you have two functions that are multiplied together and you need to differentiate the product, the Product Rule is your best friend.
It helps find the derivative of the product of two functions by providing a simple formula.
The Product Rule states:

• If you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( uv \) is given by:
• \( \frac{d}{dx}(uv) = u'v + uv' \)

In this formula:

• \( u' \) is the derivative of the first function with respect to \( x \).
• \( v' \) is the derivative of the second function with respect to \( x \).

For example, using the given functions where \( u(1) = 2 \), \( u'(1) = 0 \), \( v(1) = 5 \), and \( v'(1) = -1 \), the Product Rule helps compute the derivative at \( x = 1 \):

• \( u'(1)v(1) + u(1)v'(1) = 0 \cdot 5 + 2 \cdot (-1) = -2 \)
• Thus, the derivative of \( uv \) at \( x = 1 \) is \(-2\).

Quotient Rule

Sometimes, functions are not only multiplied but also divided, and that's where the Quotient Rule comes into play.
This rule is used to differentiate a function that is the quotient of two other functions, say \( u(x) \) over \( v(x) \).
The rule states:

• To find the derivative of \( \frac{u}{v} \), use:
• \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \)

Here:

• \( u' \) is the derivative of the numerator function, \( u(x) \).
• \( v' \) is the derivative of the denominator function, \( v(x) \).
• \( v^2 \) implies we square the denominator \( v(x) \).

In practice, using \( u(1) = 2 \), \( u'(1) = 0 \), \( v(1) = 5 \), and \( v'(1) = -1 \), calculate:

• \( \frac{0 \cdot 5 - 2 \cdot (-1)}{5^2} = \frac{2}{25} \)
• This gives the derivative value of \( \frac{u}{v} \) at \( x = 1 \) as \( \frac{2}{25} \).

Differentiable Functions

Differentiable functions are essential in calculus because they can be smoothly differentiated throughout their domain.
When a function is differentiable at a point, it means the function has a defined derivative at that point.
This derivative, essentially, is the slope of the tangent line to the function graph at that point.

• If a function \( u(x) \) is differentiable at \( x = 1 \), we can find \( u'(1) \).
• Similarly, if \( v(x) \) is differentiable, we can find \( v'(1) \).

Differentiability ensures that calculus rules, such as the Product and Quotient Rules, apply without issues.

• It allows for easy calculation of complex expressions like \( \frac{u}{v} \) or \( uv \), as the example shows.
• This is because differentiable functions are smooth, without any breaks or sharp turns, making them predictable under different calculus operations.

In short, as long as our functions are differentiable, we can find their derivatives with confidence, knowing these valuable tools are at our disposal.