Problem 39 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) is differentiable. Find an expression for the derivative of \(y\) at \(x=1\), assuming that \(f(1)=2\) and \(f^{\prime}(1)=-1\) \(y=-5 x^{3} f(x)-2 x\)

Step-by-Step Solution

Answer

The derivative of \(y\) at \(x=1\) is \(-27\).

Step 1: Identify the Components

We are given the function \( y = -5x^3f(x) - 2x \). To find its derivative, we need to differentiate each component separately: \(-5x^3f(x)\) and \(-2x\).

Step 2: Apply the Product Rule

To differentiate \(-5x^3f(x)\), we use the product rule: \(\left( u(x) \cdot v(x) \right)' = u'(x)v(x) + u(x)v'(x)\), where \(u(x) = -5x^3\) and \(v(x) = f(x)\).

Step 3: Differentiate \(u(x) = -5x^3\)

Find the derivative of \(u(x)\): \(u'(x) = -15x^2\).

Step 4: Use the Given Values for \(v(x) = f(x)\)

We know that \(f(1) = 2\) and \(f'(1) = -1\). We'll use these when calculating \( (uv)' \) at \( x = 1 \).

Step 5: Differentiate \(y\) using the product rule

Apply the product rule to \(-5x^3f(x)\), setting \( u = -5x^3 \) and \( v = f(x) \). Then \( (uv)' = -15x^2f(x) - 5x^3f'(x) \).

Step 6: Differentiate \(-2x\) Separately

The derivative of \(-2x\) is simply \(-2\).

Step 7: Combine the Derivatives

Combine the derivatives from Steps 5 and 6 to find \(y'\): \[y' = -15x^2 f(x) - 5x^3 f'(x) - 2\].

Step 8: Plug in \(x = 1\)

Substitute \(x = 1\), \(f(1) = 2\), and \(f'(1) = -1\) into the derivative:\[y'(1) = -15(1)^2(2) - 5(1)^3(-1) - 2\].

Step 9: Simplify the Expression

Calculate the expression from Step 8:- First term: \(-15 \times 2 = -30\),- Second term: \(-5 \times (-1) = 5\),- Third term: \(-2\).Thus, \[y'(1) = -30 + 5 - 2 = -27\].

Key Concepts

Product Rule

The product rule is an essential technique in calculus differentiation, specifically when you are dealing with the derivative of the product of two functions. It allows us to find the derivative of expressions where two functions are multiplied together.

Suppose you have a function defined as the product of two functions:

• \( y = u(x)v(x) \)

The product rule states that the derivative of this product is:

• \( (uv)' = u'(x)v(x) + u(x)v'(x) \)

In the provided problem, we identify our two functions as \( u(x) = -5x^3 \) and \( v(x) = f(x) \).

By applying the product rule, we differentiate each function separately and then sum their respective contributions:

• The derivative of \( u(x) \) is found first.
• Then, keep \( v(x) \) as it is.
• Finally, apply the derivative to \( v(x) \) itself while keeping \( u(x) \) constant.

Remember, this step is crucial in finding the correct derivative when functions are multiplied together.

Derivative Calculation

Calculating derivatives is at the heart of understanding how functions change, and it's often the goal when finding the rate of change at any given point. In our example, we needed to find the derivative of the function:

• \( y = -5x^3f(x) - 2x \)

To do this, we broke it down into manageable parts:

1. **Differentiate Each Component:**
Start with

• \( -5x^3f(x) \): Use the product rule here.
• For \( -2x \): Simply differentiate directly as it is a linear term.

2. Combine these derivatives to form the complete derivative of \( y \):

• \[ y' = -15x^2 f(x) - 5x^3 f'(x) - 2 \]

3. **Calculate the Specific Derivative at \( x = 1 \):**
Using the given values for the functions at this point:

• Substitute \( f(1) = 2 \) and \( f'(1) = -1 \) into the equation to find the derivative at that specific point.

This methodical approach ensures

• accuracy
• understanding of each component's contribution
• brings clarity to more complex problems.

Function Substitution

Using known values for a function and its derivative at a point facilitates efficient calculation for specific points, often a step toward deeper understanding and simplification in calculus problems.

As in this exercise, the values provided at \( x = 1 \) were:

• \( f(1) = 2 \)
• and \( f'(1) = -1 \)

Function substitution involves putting these values directly into our derived expression:

• First term in our derivative became \(-15(1)^2(2) \)
• Second term was adjusted to \(-5(1)^3(-1) \)
• Add the constant derivative term \(-2\)

This substitution turns our general derivative into a specific value at \( x = 1 \):

• \[ y'(1) = -30 + 5 - 2 = -27 \]

Function substitution thus transforms a seemingly abstract derivative into concrete results that reflect the behavior of the original function at a given point.