Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate. (a) \(y=\left(2^{2}\right)^{x}\) (b) \(y=2^{2^{x}}\) (c) \(y=\frac{e^{\pi x}}{x}\) (d) \(y=\frac{x^{3}+1}{x^{2}+1}\) (e) \(y=5 \ln \left(\frac{5 x+3}{\sqrt{x}}\right)\) (f) \(y=\frac{3}{2 \ln \left(8 x^{2}+1\right)}\)

Step-by-Step Solution

Answer

The derivatives are: \n(a) \(y' = 4x \ln(4)\) \n(b) \(y' = 2^{2^{x}+x} (\ln(2))^{2}\) \n(c) \(y' = \frac{ {\pi e^{\pi x} x - e^{\pi x}} }{x^2}\) \n(d) \(y' = \frac{ 2x^{4} +3x^{2} -2x} { (x^{2} +1)^{2}}\) \n(e) \(y' = \frac{5}{\sqrt{x}} - 15\) \n(f) \(y' = - \frac{ 48x }{ (\ln(8x^{2} + 1))^2 \cdot (8x^{2} + 1) }\)

Step 1: Differentiate \(y=\left(2^{2}\right)^{x}\)

Recall the power rule and simplify \(2^{2}\) times \(x\). The derivative is then \(y' = \ln(2^{2}) \cdot (2^{2})^x = 4x \ln(4)\).

Step 2: Differentiate \(y=2^{2^{x}}\)

Recall the chain rule. The derivative is then \(y' = 2^{2^{x}} \cdot \ln(2) \cdot 2^{x} \cdot \ln(2) = 2^{2^{x}+x} (\ln(2))^{2}\).

Step 3: Differentiate \(y=\frac{e^{\pi x}}{x}\)

Use the quotient rule and the fact that the derivative of \(e^{x}\) is \(e^{x}\). The derivative is then \(y' = \frac{ {\pi e^{\pi x} x - e^{\pi x}} }{x^2}\).

Step 4: Differentiate \(y=\frac{x^{3}+1}{x^{2}+1}\)

Again use the quotient rule. The derivative is then \(y' = \frac{ {3x^2(x^{2}+1) - 2x(x^{3}+1)} }{(x^{2} + 1)^{2}}= \frac{ 2x^{4} +3x^{2} -2x} { (x^{2} +1)^{2}}\).

Step 5: Differentiate \(y=5 \ln \left(\frac{5 x+3}{\sqrt{x}}\right)\)

Use the chain rule on the whole function, then again on the argument of the logarithm function for which we also use the quotient rule. The derivative is then \(y' = 5 \cdot \frac{1}{\frac{5x+3}{\sqrt{x}}} \cdot \left( \frac{5}{2\sqrt{x}} - \frac{5\sqrt{x} + 3\sqrt{x}}{x} \right) = \frac{5}{\sqrt{x}} - 15\).

Step 6: Differentiate \(y=\frac{3}{2 \ln \left(8 x^{2}+1\right)}\)

Apply the quotient rule and the chain rule to differentiate this function. The derivative is then \( y' = - \frac{ 3 \cdot [ \frac{1}{ \ln(8x^{2} + 1) } ] \cdot \frac{1}{8x^{2} + 1} \cdot 16x }{ 2 (\ln(8x^{2} + 1))^2 } = - \frac{ 48x }{ (\ln(8x^{2} + 1))^2 \cdot (8x^{2} + 1) }\).

Key Concepts

Power Rule

Understanding the power rule is essential when differentiating functions involving powers of x. The power rule states that if you have a function of the form \(y = x^n\), where n is a real number, the derivative of y with respect to x is \(y' = nx^{n-1}\). This rule greatly simplifies finding derivatives of polynomial functions.

For example, differentiating \(y=x^3\) gives us \(y' = 3x^2\) by applying the power rule. With practice, this rule becomes almost automatic, but it's essential to remember to multiply by the original power and then subtract one from that power to find the derivative.

Chain Rule

When you're faced with a composite function - that is, a function within another function - the chain rule is your go-to differentiation technique. It states that to find the derivative of a composite function, you differentiate the outer function, then multiply by the derivative of the inner function.

The process might sound complex but is quite logical when practiced. For instance, if we have \(y = (2x+3)^4\), applying the chain rule would involve first taking the derivative of the outer function which would give us \(4(2x+3)^3\), then multiplying by the derivative of \(2x+3\), which is 2, resulting in \(y' = 8(2x+3)^3\).

Quotient Rule

Dividing functions brings us to the quotient rule, essential when differentiating ratios of functions. The quotient rule states that the derivative of a division of two functions, \(y = \frac{u(x)}{v(x)}\), is given by \(y' = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}\).

It might come across as daunting due to the number of steps, but it can be remembered with this simple rhyme: 'low d high minus high d low, square the bottom and away we go'. Essentially, you take the bottom function times the derivative of the top function minus the top function times the derivative of the bottom function, all over the square of the bottom function.

Exponential Functions

Differentiating exponential functions requires an understanding of a new constant: \(e\), Euler's number. For any exponential function of the form \(y = e^{kx}\), where k is a constant, the derivative is simply \(y' = ke^{kx}\). This is because the base \(e\)'s rate of growth is proportional to its value, a unique property among constants.

In exponential functions with different bases like \(y = a^x\), where a is a positive constant, logarithms come into play. To differentiate such a function, the derivative is \(y' = a^x \ln(a)\), where \(\ln\) denotes the natural logarithm. Exponential functions are rampant in growth and decay problems, making their differentiation vitally important in fields such as biology and finance.

Logarithmic Differentiation

Logarithmic differentiation provides an alternative method for differentiating functions that can be difficult or unwieldy to differentiate using standard rules. It's particularly useful for functions where x is the exponent or in cases involving products or quotients of functions.

To apply logarithmic differentiation, you take the natural logarithm (ln) of both sides of the function, use the properties of logarithms to simplify, and then differentiate implicitly. When dealing with the derivative of a logarithmic function itself, remember that the derivative of \(\ln(x)\) is \(\frac{1}{x}\). This becomes invaluable when handling complex expressions, as it often simplifies multiplication into addition and division into subtraction, making the differentiation process more manageable.