Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Differentiate a. \(\quad P(t)=t^{4}+e^{2 t}\) b. \(\quad P(t)=t^{4} e^{2 t}\) c. \(\quad P(t)=\frac{t^{4}}{e^{2 t}}\) d. \(P(t)=\left(e^{2 t}\right)^{4}\) e. \(P(t)=e^{2 t^{4}}\) f. \(P(t)=\left(t^{2}+1\right)^{4}(5 t+1)^{7}\) g. \(\quad P(t)=(\ln t)^{3}\) h. \(\quad P(t)=\left(e^{3 t} \ln 2 t\right)^{4}\) i. \(\quad P(t)=\frac{\ln t}{t}\) j. \(\quad P(t)=t \ln t-t\) k. \(P(t)=\frac{1}{\ln t}\) l. \(P(t)=e^{(\ln t)}\) m. \(P(t)=\frac{5 t^{2}-2 t-7}{t^{2}+1}\) n. \(P(t)=\frac{(t+2)^{2}}{t^{2}+2}\)

Step-by-Step Solution

Answer

Differentiate using rules like power, product, quotient, and chain for each expression.

Step 1: Differentiate Using Power and Exponential Rules (a)

For the function \( P(t) = t^4 + e^{2t} \), we use the power rule \( \frac{d}{dt}[t^n] = nt^{n-1} \) and the rule for differentiating exponential functions \( \frac{d}{dt}[e^{kt}] = ke^{kt} \). Thus:\[ \frac{d}{dt}[t^4 + e^{2t}] = 4t^3 + 2e^{2t}. \]

Step 2: Apply Product Rule (b)

The function is \( P(t) = t^4 e^{2t} \). The product rule states \( \frac{d}{dt}[uv]=u'v+uv' \). Let \( u = t^4 \) and \( v = e^{2t} \). Then:\\( u' = 4t^3 \), and \( v' = 2e^{2t} \).Thus, \( \frac{d}{dt}[t^4 e^{2t}] = 4t^3 e^{2t} + t^4(2 e^{2t}) = t^3 e^{2t} (4 + 2t) \).

Step 3: Use Quotient Rule (c)

For \( P(t) = \frac{t^4}{e^{2t}} \), use the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Let \( u = t^4 \), \( u' = 4t^3 \), \( v = e^{2t} \), and \( v' = 2e^{2t} \).The derivative is:\[ \frac{4t^3 e^{2t} - t^4(2 e^{2t})}{(e^{2t})^2} = \frac{2t^3 e^{2t} (2 - t)}{e^{4t}}. \]

Step 4: Apply Chain Rule for Exponential Functions (d)

The function \( P(t) = (e^{2t})^4 \) can be rewritten as \( e^{8t} \). Then differentiate using the exponential rule:\[ \frac{d}{dt}[e^{8t}] = 8e^{8t}. \]

Step 5: Utilize Chain Rule for Composite Functions (e)

For \( P(t) = e^{2t^4} \), apply the chain rule:Let \( f(u) = e^u \) with \( u = 2t^4 \), then \( \frac{du}{dt} = 8t^3 \).The derivative is \[ \frac{d}{dt}[e^{2t^4}] = e^{2t^4} \cdot 8t^3. \]

Step 6: Use Product Rule and Chain Rule (f)

For \( P(t) = (t^2+1)^4(5t+1)^7 \), apply the product rule: \( \frac{d}{dt}[uv] = u'v + uv' \) and the chain rule. Let \( u = (t^2+1)^4 \), \( u' = 4(t^2+1)^3 \, \cdot 2t \). Let \( v = (5t+1)^7 \), \( v' = 7(5t+1)^6 \, \cdot 5 \).Then:\[ \frac{d}{dt}[(t^2+1)^4(5t+1)^7] = 4(t^2+1)^3(2t)(5t+1)^7 + (t^2+1)^4 \cdot 7(5t+1)^6 \cdot 5. \]

Step 7: Apply Chain Rule for Logarithm (g)

For \( P(t) = (\ln t)^3 \), use the chain rule:\[ \frac{d}{dt}[(\ln t)^3] = 3(\ln t)^2 \cdot \frac{1}{t}. \]

Step 8: Use Chain Rule for Complex Exponential Log Functions (h)

For \( P(t) = (e^{3t} \ln 2t)^4 \), apply the chain rule. Set \( u = (e^{3t} \ln 2t) \). Then:\[ \frac{du}{dt} = e^{3t}(3 \ln 2t + \frac{1}{2t}), \]\[ \frac{d}{dt}[(e^{3t} \ln 2t)^4] = 4(e^{3t} \ln 2t)^3 \, \cdot e^{3t} \left( 3 \ln 2t + \frac{1}{2t} \right). \]

Step 9: Combine Quotient and Logarithm Rules (i)

For \( P(t) = \frac{\ln t}{t} \), apply the quotient rule:Let \( u = \ln t \), \( u' = \frac{1}{t} \), \( v = t \), \( v' = 1 \).Thus, \[ \frac{d}{dt}[\frac{\ln t}{t}] = \frac{\frac{1}{t}t - \ln t \cdot 1}{t^2} = \frac{1 - \ln t}{t^2}. \]

Step 10: Differentiate using Product and Logarithm Rules (j)

For \( P(t) = t \ln t - t \), use both the product rule and simple derivation:\[ \frac{d}{dt}[t \ln t - t] = \left(1 \cdot \ln t + t \cdot \frac{1}{t}\right) - 1 = \ln t + 1 - 1 = \ln t. \]

Step 11: Use Chain Rule for Logarithm Function (k)

For \( P(t) = \frac{1}{\ln t} \), rewrite as \( (\ln t)^{-1} \) and use the chain rule:\[ \frac{d}{dt}[\ln t]^{-1} = -1(\ln t)^{-2} \cdot \frac{1}{t} = -\frac{1}{t(\ln t)^2}. \]

Step 12: Simplify Exponential Expression (l)

For \( P(t) = e^{\ln t} \), simplify using the property of exponents \((e^{\ln x} = x)\):Thus, \( P(t) = t \), then differentiate \( \frac{d}{dt}[t] = 1 \).

Step 13: Apply Quotient Rule (m)

For \( P(t) = \frac{5t^2 - 2t - 7}{t^2 + 1} \), use the quotient rule:Let \( u = 5t^2 - 2t - 7 \), \( u' = 10t - 2 \), \( v = t^2 + 1 \), \( v' = 2t \).The derivative is:\[ \frac{10t(t^2 + 1) - (5t^2 - 2t - 7)(2t)}{(t^2 + 1)^2}. \]

Step 14: Apply Quotient Rule Again (n)

For \( P(t) = \frac{(t+2)^2}{t^2+2} \), use the quotient rule:Let \( u = (t+2)^2 \), \( u' = 2(t+2) \), \( v = t^2+2 \), \( v' = 2t \).The derivative is:\[ \frac{2(t+2)(t^2+2) - (t+2)^2(2t)}{(t^2+2)^2}. \]

Key Concepts

Product Rule

When differentiating a function where two or more functions are multiplied together, the product rule comes in handy. The basic formula for the product rule is \( \frac{d}{dt}[uv] = u'v + uv' \), where \( u \) and \( v \) are functions of \( t \). This rule is useful when tackling problems like \( P(t) = t^4 e^{2t} \).
**How It Works:**
1. Identify the two functions that are being multiplied. For example, in \( P(t) = t^4 e^{2t} \), \( u = t^4 \) and \( v = e^{2t} \).
2. Differentiate \( u \) to find \( u' \) and differentiate \( v \) to get \( v' \). Here, \( u' = 4t^3 \) and \( v' = 2e^{2t} \).
3. Apply the product rule formula: \( \frac{d}{dt}[t^4 e^{2t}] = u'v + uv' = 4t^3 e^{2t} + 2t^4 e^{2t} \).
4. Simplify if possible. In this case, factor out common terms: \( t^3 e^{2t} (4 + 2t) \).
The product rule is particularly useful for functions wherein both parts change with respect to the variable and need to be considered separately. Remember, it's all about multiplying and adding the differential parts together.

Quotient Rule

Differentiating a function that involves division of two other functions requires the quotient rule. This rule is expressed as \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). It can wonderfully simplify the task of differentiating rational expressions like \( P(t) = \frac{t^4}{e^{2t}} \).
**Steps to Apply the Quotient Rule:**
- First, identify \( u \) and \( v \). For instance, with \( P(t) = \frac{t^4}{e^{2t}} \), \( u = t^4 \) and \( v = e^{2t} \).
- Differentiate \( u \) to find \( u' \), and \( v \) to achieve \( v' \). Hence, \( u' = 4t^3 \) and \( v' = 2e^{2t} \).
- Plug into the formula: \( \frac{d}{dt}\left[\frac{t^4}{e^{2t}}\right] = \frac{(4t^3)(e^{2t}) - (t^4)(2e^{2t})}{(e^{2t})^2} \).
- Simplify: This results in \( \frac{2t^3 e^{2t} (2 - t)}{e^{4t}} \). Factoring and reducing can help simplify the expression further.
The quotient rule is perfect for cases where a variable expression in the numerator is being divided by another in the denominator. Consider both parts of the division equally; this balances their differential contributions.

Chain Rule

The chain rule is used when differentiating composite functions. A function is termed composite when one function is nested inside another, and the chain rule helps manage the intricate layer of differentiation required. For example, with a function like \( P(t) = e^{2t^4} \), the chain rule is particularly useful.
**Implementing the Chain Rule:**
- Identify the outer function and the inner function. With \( P(t) = e^{2t^4} \), the outer function is \( e^u \), where \( u = 2t^4 \).
- Differentiate \( u \), to get \( du/dt = 8t^3 \).
- Then, differentiate the outer function \( e^u \) with respect to \( u \) itself, not \( t \).
- Multiply the derivative of the outer function with the derivative of the inner function: \( \frac{d}{dt}[e^{2t^4}] = e^{2t^4} \cdot 8t^3 \).
Use the chain rule when the function is wrapped inside another, with a variable that needs chaining through both derivatives. It essentially "links together" the derivatives, ensuring the inner function's effect is properly scaled and accounted for. Owing to its versatility, it’s a frequently employed tool in calculus for intricate function compositions.