Problem 12 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(f(x)=\left(3 x^{2}-1\right) e^{1-x^{2}}\)

Step-by-Step Solution

Answer

The derivative is \(f'(x) = (-6x^3 + 8x)e^{1-x^2}\).

Step 1: Identify Differentiation Rule

We need to differentiate the function \(f(x) = (3x^2 - 1)e^{1-x^2}\). This function is a product of two functions: \(u(x) = 3x^2 - 1\) and \(v(x) = e^{1-x^2}\). We will use the product rule for differentiation, which states: if \(f(x) = u(x)v(x)\), then \(f'(x) = u'(x)v(x) + u(x)v'(x)\).

Step 2: Differentiate the First Function

Differentiate \(u(x) = 3x^2 - 1\) with respect to \(x\). The derivative is \(u'(x) = 6x\) since the derivative of \(3x^2\) is \(6x\) and the derivative of a constant is \(0\).

Step 3: Differentiate the Second Function

Differentiate \(v(x) = e^{1-x^2}\) with respect to \(x\). This requires the chain rule. The outer function is \(e^u\) where \(u = 1 - x^2\). The derivative of the outer function is \(e^u\) and the derivative of the inner function \(1-x^2\) is \(-2x\). Therefore, \(v'(x) = -2xe^{1-x^2}\).

Step 4: Apply the Product Rule

Using the product rule, \(f'(x) = u'(x)v(x) + u(x)v'(x)\). Substitute the derivatives we calculated:\[f'(x) = (6x)(e^{1-x^2}) + (3x^2 - 1)(-2xe^{1-x^2})\]

Step 5: Simplify the Expression

Simplify the expression obtained in the previous step:\[f'(x) = 6xe^{1-x^2} - 2x(3x^2 - 1)e^{1-x^2}\]. Distribute and combine like terms:\[f'(x) = 6xe^{1-x^2} - (6x^3 - 2x)e^{1-x^2} = (6x - 6x^3 + 2x)e^{1-x^2}\]. Combine the linear terms:\[f'(x) = (-6x^3 + 8x)e^{1-x^2}\].

Key Concepts

Product Rule

When dealing with problems involving differentiation, the product rule is essential whenever you encounter a function that results from multiplying two differentiable functions. The product rule can be represented as follows: if you have a function \( f(x) = u(x) \, v(x) \), where \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative \( f'(x) \) is given by:

• \( f'(x) = u'(x) \, v(x) + u(x) \, v'(x) \)

This rule helps us find the rate of change of the product of two functions. Imagine you are given a function \( f(x) = (3x^2 - 1) \, e^{1-x^2} \). The components are \( u(x) = 3x^2 - 1 \) and \( v(x) = e^{1-x^2} \). Each of these components needs to be differentiated separately before we can combine them using the product rule. This differentiation enables us to understand how each part of the function contributes to the overall rate of change. Breaking down the calculation into simple steps helps avoid mistakes and ensures our understanding of how individual components affect the whole function.

Chain Rule

The chain rule comes into play when you need to differentiate composite functions. A composite function is when one function is "inside" another function. In our example, we look at the expression \( v(x) = e^{1-x^2} \), which is a composition of functions. We can see it as an outer function \( e^u \) and an inner function \( u = 1-x^2 \). The chain rule allows us to differentiate such compositions effectively.Using the chain rule for a composed function \( y = g(u(x)) \), the derivative \( y'(x) \) is determined by:

• \( y'(x) = g'(u(x)) \cdot u'(x) \)

This means we differentiate the outer function, leaving the inner function unchanged, and then multiply it by the derivative of the inner function. So for our function \( v(x) \), we differentiate:

• The outer \( e^u \), which remains \( e^{1-x^2} \)
• The inner \( 1-x^2 \), which becomes \(-2x\)

Combining these using the chain rule gives \( v'(x) = -2x \, e^{1-x^2} \). This showcases how changes in \( x \) influence both the inner and outer functions.

Derivative

Derivatives are fundamental in calculus as they provide a way to measure how a function changes as its input changes. Essentially, the derivative of a function at a certain point gives the slope of the tangent line to the function's graph at that point. With derivatives, we capture the notion of instantaneous rate of change.When working through the solution steps, first we found \( u'(x) \), the derivative of \( u(x) = 3x^2 - 1 \) which is \( 6x \). This highlights how the quadratic component of the function changes at each point \( x \). Next, we used the chain rule to find \( v'(x) \), which shows how complex exponential functions behave.Putting it all together with the product rule, these derivatives help us find the overall rate of change of the product function. In simplified terms, after applying the product rule, the function's derivative is \( f'(x) = (-6x^3 + 8x) \, e^{1-x^2} \). This result captures the combined influence of linear, quadratic, and exponential changes in a neat expression. Understanding derivatives allows us to make sense of behavior across various mathematical contexts.