Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate the following, simplifying the expression first if useful. (a) \(y=\pi e^{3 t^{2}+\pi}\) (b) \(y=\ln \left(e^{t}+1\right)\) (c) \(y=\frac{\pi^{2}}{\sqrt{x^{2}+4}}\) (d) \(y=\frac{1}{(\ln x)^{2.6}}\) (e) \(y=\frac{1}{\left(\ln x^{2}\right)^{1.5}}\) (f) \(y=\sqrt[3]{\ln \left(e^{t}+1\right)}\)

Step-by-Step Solution

Answer

The derivatives are (a) \(y' = 6\pi te^{3 t^{2}+\pi}\), (b) \(y'=\frac{e^t}{e^t + 1}\), (c) \(y' = -\frac{\pi^2 x}{(x^2 + 4)^{3/2}}\), (d) \(y' = -\frac{2.6}{x (\ln x)^{3.6}}\), (e) \(y'=\frac{-1.5x}{(\ln x)^{2.5}}\), and (f) \(y'=\frac{e^t}{3(\ln(e^t + 1))^{2/3}(e^t + 1)}\).

Step 1: Differentiating \(y=\pi e^{3 t^{2}+\pi}\)

Use the chain rule formula for differentiation of \(\exp(f(x))\), which is \(\exp(f(x)) \cdot f'(x)\). Here \(f(t) = 3t^2 + \pi\). So the derivative is \(y' = \pi e^{3 t^{2}+\pi} \cdot (6t)\)

Step 2: Differentiating \(y=\ln \left(e^{t}+1\right)\)

Here, use the chain rule for differentiation of \(\ln(u)\), which is \(u'/u\). With \(u = e^t + 1\), the derivative is \(y' = \frac{e^t}{e^t + 1}\)

Step 3: Differentiating \(y=\frac{\pi^{2}}{\sqrt{x^{2}+4}}\)

Use the quotient rule and chain rule for differentiation. The quotient rule for \(f/g\) is \((f'g - fg')/g^2\) and chain rule for \(\sqrt{u}\) is \(0.5u^{-0.5}u'\). So, \(y' = -\frac{\pi^2 x}{(x^2 + 4)^{3/2}}\)

Step 4: Differentiating \(y=\frac{1}{(\ln x)^{2.6}}\)

Use the chain rule and the rule for differentiating the natural logarithm here. The derivative is \(y' = -\frac{2.6}{x (\ln x)^{3.6}}\)

Step 5: Differentiating \(y=\frac{1}{\left(\ln x^{2}\right)^{1.5}}\)

Again use the chain rule and the rule for differentiating the natural logarithm. Here, \(y' = -\frac{1.5x}{(\ln x)^{2.5}}\)

Step 6: Differentiating \(y=\sqrt[3]{\ln \left(e^{t}+1\right)}\)

Use the chain rule and the rule for differentiating the cube root and natural logarithm. Here, \(y'=\frac{e^t}{3(\ln(e^t + 1))^{2/3}(e^t + 1)}\).

Key Concepts

Chain Rule

To understand the chain rule, imagine you're peeling multiple layers to get to the core. In calculus, the chain rule is essential when you have a function nested within another function. It helps us differentiate composites of functions easily.

Consider the differentiation of the function \( y = \pi e^{3t^2 + \pi} \). Here, we have an outer exponential function and an inner function \( f(t) = 3t^2 + \pi \). The chain rule in general form is:

• If \( y = u(v(x)) \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \).
• For \( y = \pi e^{3t^2+\pi} \), we let \( u = f(t) = 3t^2+\pi \).
• The derivative becomes \( y' = \pi e^{3t^2+\pi} \cdot 6t \).

The chain rule allows us to focus on differentiating the inside function first and applying the result to the outside function.

Quotient Rule

The quotient rule is indispensable when dealing with ratios of two functions. It's like applying a special formula that keeps it neat and organized. You only need to remember the rule and apply it systematically.

The formula for differentiating \( \frac{f(x)}{g(x)} \) is:

• \( \left(\frac{f}{g}\right)' = \frac{f' \cdot g - f \cdot g'}{g^2} \)

For the function \( y = \frac{\pi^2}{\sqrt{x^2 + 4}} \), we identify:

• \( f = \pi^2 \) and \( g = \sqrt{x^2 + 4} \).
• The derivative of \( g \) using the chain rule is \( 0.5(x^2 + 4)^{-0.5} \cdot 2x = \frac{x}{\sqrt{x^2+4}} \).
• Plug these into the quotient rule to get \( y' = -\frac{\pi^2 x}{(x^2 + 4)^{3/2}} \).

Exponential Functions

Exponential functions are the powerhouses of natural growth and decay. They involve constants like \( e \), the base of the natural logarithm, which is about 2.718.

An exponential function can be of the form \( y = e^{f(x)} \). Their differentiation relies heavily on the chain rule.

• When differentiating \( y = e^{3t^2+\pi} \), recognize that \( e^{f(t)} \) keeps its form, \( e^{3t^2+\pi} \).
• Multiply by the derivative of the exponent, resulting in \( y' = e^{3t^2+\pi} \cdot 6t \) as the chain rule dictates.

Exponential functions grow very quickly or slowly, depending on whether the exponent is positive or negative.

Logarithmic Differentiation

Logarithmic differentiation can transform complex expressions into simpler ones. It's especially useful when dealing with products, quotients, or powers raised to a variable.

Consider the function \( y = \ln(e^t + 1) \). To differentiate, use the basic property of logarithms and the chain rule:

• The derivative formula \( \frac{d}{dx} \ln(u) = \frac{u'}{u} \) is applied.
• For \( \ln(e^t + 1) \), \( u = e^t + 1 \) and \( u' = e^t \).
• Thus, \( y' = \frac{e^t}{e^t + 1} \).

If you have a logarithmic function combined with others, logarithmic differentiation helps to break it down into simpler derivatives.