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 Function and Its Components

We are given the function \( y = -5x^3 f(x) - 2x \) and need to find its derivative evaluated at \( x = 1 \). This function consists of two terms: \( -5x^3 f(x) \) and \( -2x \). Each has its own derivative rule.

Step 2: Differentiate Each Term

To find the derivative of \( y \) with respect to \( x \), differentiate each part of the function separately. The derivative of \(-5x^3 f(x)\) is found using the product rule. The derivative of \(-2x\) is straightforward.

Step 3: Apply the Product Rule

For the term \(-5x^3 f(x)\), apply the product rule: if \( u = -5x^3 \) and \( v = f(x) \), then \( \frac{d}{dx}[uv] = u'v + uv' \). This yields: \[ \frac{d}{dx}[-5x^3 f(x)] = (-15x^2)f(x) + (-5x^3)f'(x). \]

Step 4: Differentiate the Linear Term

The derivative of \(-2x\) is \(-2\) since the derivative of \(x\) is \(1\) and the derivative of a constant is zero.

Step 5: Combine the Derivatives

Combine the results from Steps 3 and 4 to find the total derivative of \( y \). The derivative is:\[ y' = (-15x^2)f(x) + (-5x^3)f'(x) - 2. \]

Step 6: Evaluate the Derivative at \( x = 1 \)

Substitute \( x = 1 \), \( f(1) = 2 \), and \( f'(1) = -1 \) into the derivative:\[ y' = (-15(1)^2)(2) + (-5(1)^3)(-1) - 2. \]Simplify this to get:\[ y' = (-30) + (5) - 2 = -27. \]

Key Concepts

Understanding the Product Rule

In calculus, the product rule is a crucial tool used to differentiate functions that are products of two or more functions. If you have a function which is the product of two functions, like \(u(x)\) and \(v(x)\), and you need to differentiate it, the product rule comes into play.

• According to the product rule, if \( y = u(x)v(x) \), then the derivative \( y' \) is given by \( u'(x)v(x) + u(x)v'(x) \).
• This means you first differentiate \(u(x)\), multiply it by \(v(x)\), and then add it to the product of \(u(x)\) and the derivative of \(v(x)\).

This is precisely what is done in our step where the derivative of \(-5x^3 f(x)\) is computed. Here, \(u(x) = -5x^3\) and \(v(x) = f(x)\). Thus:

• The derivative of \(-5x^3\) is \(-15x^2\).
• The function \(f(x)\) remains undifferentiated initially.
• Then you add \(-5x^3\) multiplied by the derivative of \(f(x)\), denoted as \(f'(x)\).

Combining these, we get the derivative of the term as \((-15x^2)f(x) + (-5x^3)f'(x)\). This powerful rule allows us to handle more complex expressions with ease.

Basics of Differentiation

Differentiation is the fundamental concept of calculus that deals with finding the derivative of a function. The derivative measures how a function changes as its input changes.

• The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function.
• The derivative of the basic function \(x\) is \(1\). Thus the derivative of a constant term multiplied by \(x\), like \(-2x\), is straightforwardly \(-2\).

Combining differentiation rules with algebraic operations, we can handle more complicated functions.
For example, in our function \(y = -5x^3 f(x) - 2x\), the term \(-2x\) is linear and its derivative is simply \(-2\). Differentiation is a tool that enables us to understand the rates of change within functions, forming the foundation for evaluating derivatives.

Evaluating the Derivative at a Specific Point

Once you have determined the derivative of a function, the next step often involves evaluating this derivative at a specific point to understand the behavior of the function at that point.
For instance, in our exercise, the function \(y = -5x^3 f(x) - 2x\) has its derivative calculated as \[ y' = (-15x^2)f(x) + (-5x^3)f'(x) - 2. \] Now, to evaluate this at \(x = 1\):

• Substitute \(x = 1\) into the derivative equation.
• Use the provided values \(f(1) = 2\) and \(f'(1) = -1\).

Substituting these into the equation: \[ y' = (-15(1)^2)(2) + (-5(1)^3)(-1) - 2. \] After simplifying, the derivative evaluates to \(y' = -27\). This evaluated derivative tells us the rate of change of \(y\) specifically at the point where \(x = 1\), providing valuable insight into the function's behavior at this point.