Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

We introduced the power chain rule \(\left[(u(x))^{n}\right]^{\prime}=n(u(x))^{n-1}[u(x)]^{\prime}\) for fractional and negative exponents, \(n\), in Section 4.3 .1 (see Exercises 4.3 .3 and 4.3 .4 ). Use these rules when necessary in the following exercise. Compute \(y^{\prime}(x)\) and \(y^{\prime \prime}(x)\) for a. \(y(x)=x^{2}+e^{x}\) b. \(y(x)=3 x^{2}+2 e^{x}\) c. \(y(x)=\left(1+e^{x}\right)^{2}\) d. \(y(x)=\left(e^{x}\right)^{2}\) e. \(y(x)=e^{2 x}=\left(e^{x}\right)^{2}\) f. \(y(x)=e^{-x} \quad=\left(e^{x}\right)^{-1}\) g. \(y(x)=e^{3 x} \quad=\left(e^{x}\right)^{3}\) h. \(y(x)=e^{x} \times e^{2 x}\) i. \(y(x)=\left(5+e^{x}\right)^{3}\) j. \(y(x)=\frac{1}{1+e^{x}} \quad=\left(1+e^{x}\right)^{-1}\) k. \(y(x)=\sqrt{e^{x}} \quad=\left(e^{x}\right)^{\frac{1}{2}}\) l. \(y(x)=e^{\frac{1}{2} x}\) m. \(y(x)=e^{0.6 x}=\left(e^{x}\right)^{0.6}\) n. \(y(x)=e^{-0.005 x}\)

Step-by-Step Solution

Answer

Use derivatives rules and power chain rule for each function.

Step 1: Exercise a - Calculate first derivative

Given function: \(y(x) = x^2 + e^x\). We find the first derivative using the sum rule for derivatives. For \(x^2\), use \(\frac{d}{dx}[x^2] = 2x\). For \(e^x\), use \(\frac{d}{dx}[e^x] = e^x\). Thus, the derivative is \(y'(x) = 2x + e^x\).

Step 2: Exercise a - Calculate second derivative

Differentiating \(y'(x) = 2x + e^x\) again, we apply the derivative rules. For \(2x\), \(\frac{d}{dx}[2x] = 2\). For \(e^x\), \(\frac{d}{dx}[e^x] = e^x\). Thus, the second derivative is \(y''(x) = 2 + e^x\).

Step 3: Exercise b - First and second derivatives

Given \(y(x) = 3x^2 + 2e^x\). First derivative: \(y'(x) = 6x + 2e^x\). Second derivative: \(y''(x) = 6 + 2e^x\).

Step 4: Exercise c - First and second derivatives using chain rule

Given \(y(x) = (1 + e^x)^2\). Let \(u(x) = 1 + e^x\), then \(y = u^2\). \(y'(x) = 2u \cdot u'(x) = 2(1 + e^x) \cdot e^x\). \(y''(x)\) involves product rule: \(y''(x) = 2(e^x + e^{2x})\).

Step 5: Exercise d - First and second derivatives using the identity

Given \(y(x) = (e^x)^2\). The identity \((e^x)^2 = e^{2x}\) implies \(y'(x) = 2e^{2x}\) and \(y''(x) = 4e^{2x}\).

Step 6: Exercise e - Recalculate \(y(x) = e^{2x}\) using the identity

Same function as in (d), \(y'(x) = 2e^{2x}\) and \(y''(x) = 4e^{2x}\).

Step 7: Exercise f - Derivatives for a negative exponent

Given \(y(x) = e^{-x}\), differentiating gives \(y'(x) = -e^{-x}\) and \(y''(x) = e^{-x}\).

Step 8: Exercise g - First and second derivatives for power rule

Given \(y(x) = e^{3x}\), first derivative is \(y'(x) = 3e^{3x}\) and the second derivative is \(y''(x) = 9e^{3x}\).

Step 9: Exercise h - Derivatives using product rule

Given \(y(x) = e^x \cdot e^{2x}\), we use the identity \(e^{x+2x}=e^{3x}\). Thus, \(y'(x) = 3e^{3x}\) and \(y''(x) = 9e^{3x}\).

Step 10: Exercise i - Derivatives using chain rule

Given \(y(x) = (5+e^x)^3\), let \(u(x) = 5+e^x\). Then use \((u(x))^3\) chain rule: \(y'(x) = 3(5+e^x)^2 \cdot e^x\). For \(y''(x)\), apply product rule: \(y''(x) = 3[(10e^x + e^{2x}(5+e^x))]\).

Step 11: Exercise j - Multiply by standard derivative for negative power

Given \(y(x) = \frac{1}{1+e^x}\), use \(y(x) = (1+e^x)^{-1}\). Differentiating, \(y'(x) = -(1+e^x)^{-2}e^x\). The second derivative involves applying the chain and product rules.

Step 12: Exercise k - Derivatives for square root

Given \(y(x) = \sqrt{e^x}\), use \((e^x)^{1/2}\). Thus, \(y'(x) = \frac{1}{2}(e^{x})^{-1/2}e^x\). Simplify to find \(y''(x)\) using derivative rules for fractional powers.

Step 13: Exercise l - Derivatives for a constant coefficient

Given \(y(x) = e^{\frac{1}{2}x}\), apply standard rules: \(y'(x) = \frac{1}{2}e^{\frac{1}{2}x}\), \(y''(x) = \frac{1}{4}e^{\frac{1}{2}x}\).

Step 14: Exercise m - Derivatives for decimal exponent

For \(y(x) = e^{0.6x}\), first derivative is \(y'(x) = 0.6e^{0.6x}\), and the second derivative is \(y''(x) = 0.36e^{0.6x}\).

Step 15: Exercise n - Negative constant exponent derivatives

Given \(y(x) = e^{-0.005x}\), first derivative is \(y'(x) = -0.005e^{-0.005x}\) and second derivative is \(y''(x) = 0.000025e^{-0.005x}\).

Key Concepts

Power Chain Rule

When dealing with functions raised to a power, the Power Chain Rule is your go-to tool for differentiation. This rule is particularly handy for functions of the form \((u(x))^n\), where \(u(x)\) is any function of \(x\) and \(n\) is a real number.

• The rule states: \([u(x)]^n]' = n[u(x)]^{n-1} \cdot u'(x)\).
• Essentially, you bring down the power as a coefficient and then reduce the power by one.

You also need to multiply by the derivative of the inner function, \(u'(x)\). This is like peeling away an outer layer (the power) while considering what lies inside (the function). This approach is powerful when facing fractional or negative powers, making complex differentiation a bit simpler. By applying this rule, you ensure the process is straightforward, even when dealing with more challenging functions.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. A pure form of exponential is \(e^x\), where "e" is Euler's number, a constant approximately equal to 2.71828.

• The derivative of \(e^x\) is unique because it remains the same: \(\frac{d}{dx}[e^x] = e^x\).
• This property simplifies calculations considerably.

However, exponential functions can become more complex when the exponent is a function of \(x\), say, \(e^{g(x)}\). Here, you'd apply the chain rule to find the derivative: \(\frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x)\).In real-world applications, exponential functions describe growth and decay processes, such as population growth or radioactive decay.

Product Rule

The Product Rule is crucial when differentiating products of two or more functions. It allows you to determine how the rate of change of a product is related to the rates of change of its individual factors. The Product Rule can be remembered by the mnemonic: "first times the derivative of the second plus second times the derivative of the first."

• If \(y(x) = u(x) \, v(x)\), then \(y'(x) = u(x) \, v'(x) + v(x) \, u'(x)\).
• This helps in unraveling the complexity involved when functions are intertwined through multiplication.

You apply the rule sequentially, ensuring each function and its partner's derivative are accounted for. This method preserves the mathematical harmony of the function product while unraveling its individual changes over time.

Sum Rule

The Sum Rule of differentiation simplifies the process when dealing with the sum of multiple functions. Essentially, it states that the derivative of a sum is the sum of the derivatives.

• For functions \(y(x) = u(x) + v(x)\), the derivative is \(y'(x) = u'(x) + v'(x)\).
• This rule breaks down larger problems into manageable pieces, ensuring accuracy and simplicity.

By applying the Sum Rule, you can easily compute the derivative of complex expressions where each part might have different mathematical characteristics. It is particularly useful when working with polynomials, where each term may behave differently but combines to form a cohesive whole. By treating each function individually before summing their results, you maintain clarity and precision.