Question

Find the derivatives of: $$\text { (a) } y=\ln \left(7 t^{5}\right)$$ $$\text { (b) } y=\ln \left(a t^{c}\right)$$ $$\text { (c) } y=\ln (t+19)$$ $$\text { (d) } y=5 \ln (t+1)^{2}$$ $$\text { (e) } y=\ln x-\ln (1+x)$$ $$\text { (f) } y=\ln \left[x(1-x)^{8}\right]$$ $$\text { (g) } y=\ln \left(\frac{2 x}{1+x}\right)$$ $$\text { (h) } y=5 x^{4} \ln x^{2}$$

Short answer

The derivatives are: (a) \( \frac{5}{t} \), (b) \( \frac{c}{t} \), (c) \( \frac{1}{t+19} \), (d) \( \frac{10}{t+1} \), (e) \( \frac{1}{x} - \frac{1}{1+x} \), (f) \( \frac{1}{x} - \frac{8}{1-x} \), (g) \( \frac{1}{x} - \frac{1}{1+x} \), (h) \( 10x^3 + 40x^3 \ln(x) \).

Step-by-step solution

Differentiate y=ln(7t^5) using the chain rule

To differentiate the function \( y = \ln(7t^5) \), apply the chain rule. The derivative of \( \ln(u) \), where \( u = 7t^5 \), is \( \frac{1}{u} \cdot \frac{du}{dt} \). First, find \( \frac{du}{dt} \) where \( u = 7t^5 \): the derivative is \( 35t^4 \). Thus, the derivative of \( y \) is \[ \frac{dy}{dt} = \frac{1}{7t^5} \times 35t^4 = \frac{5}{t}. \]

Differentiate y=ln(a t^c) using properties of logarithms

For \( y = \ln(a t^c) \), apply the properties of logarithms: \( \ln(a t^c) = \ln(a) + c \ln(t) \). The derivative of a constant \( \ln(a) \) is zero, and the derivative of \( c \ln(t) \) with respect to \( t \) is \( \frac{c}{t} \). Therefore, \( \frac{dy}{dt} = \frac{c}{t}. \)

Differentiate y=ln(t+19) using the chain rule

Set \( u = t + 19 \) so \( y = \ln(u) \). The derivative \( \frac{du}{dt} = 1 \). Using the formula \( \frac{d}{dt} \ln(u) = \frac{1}{u} \cdot \frac{du}{dt} \), we find \( \frac{dy}{dt} = \frac{1}{t+19}. \)

Differentiate y=5ln(t+1)^2 using properties and chain rule

Rewrite \( y = 5 \ln((t+1)^2) = 10 \ln(t+1) \) using \( \ln(a^b) = b \ln(a) \). Then, differentiate \( 10 \ln(t+1) \) to get \( \frac{10}{t+1}. \)

Differentiate y=ln(x)-ln(1+x) using properties of logarithms

Use the property \( \ln \left(\frac{a}{b}\right) = \ln(a) - \ln(b) \). The derivative of \( \ln(x) \) is \( \frac{1}{x} \), and the derivative of \( \ln(1+x) \) is \( \frac{1}{1+x} \). Therefore, \( \frac{dy}{dx} = \frac{1}{x} - \frac{1}{1+x}. \)

Differentiate y=ln[x(1-x)^8] using derivatives of product

Express \( y = \ln(x) + 8 \ln(1-x) \). The derivative of \( \ln(x) \) is \( \frac{1}{x} \), and the derivative of \( \ln(1-x) \) is \( -\frac{1}{1-x} \). Combine to get \( \frac{dy}{dx} = \frac{1}{x} - \frac{8}{1-x}. \)

Differentiate y=ln(2x/1+x) using quotient rule

The expression simplifies to \( \ln(2x) - \ln(1+x) \). Differentiate \( \frac{d}{dx} \ln(2x) = \frac{1}{x} \) and \( \frac{d}{dx} \ln(1+x) = \frac{1}{1+x} \). Therefore, \( \frac{dy}{dx} = \frac{1}{x} - \frac{1}{1+x}. \)

Differentiate y=5x^4ln(x^2) using product and chain rule

For \( y = 5x^4 \ln(x^2) = 10 x^4 \ln(x) \), use the product rule. The derivative of \( 10x^4 \) is \( 40x^3 \), and the derivative of \( \ln(x) \) is \( \frac{1}{x} \). Thus, for the product, \[ \frac{dy}{dx} = 10x^4 \cdot \frac{1}{x} + 40x^3 \ln(x) = 10x^3 + 40x^3 \ln(x). \]

Key concepts

Derivatives

A derivative represents how a function changes as its input changes. In calculus, derivatives are used to determine the rate of change or the slope of a curve at a particular point. When you differentiate a function, you are essentially finding the derivative.

• If the function is a power function like \( t^n \), the derivative is found using the power rule: the derivative of \( t^n \) is \( nt^{n-1} \).
• For exponential functions, and natural logarithms, the derivatives have specific rules, such as the derivative of \( e^x \) is \( e^x \) and for \( \ln(x) \), it's \( \frac{1}{x} \).

Understanding how derivatives work allows for more complex calculations, particularly when multiple rules like the product or chain rule need to be applied. It's crucial to always simplify expressions, like in the solutions for the derivatives of logarithmic functions.

Chain Rule

The chain rule is an essential tool in calculus for finding derivatives of composite functions. A composite function is essentially a "function within a function." The chain rule allows us to differentiate these layered functions.
For example, if a function \( y \) is described by another function \( u \) such that \( y = f(g(t)) \), you use the chain rule. According to the chain rule, the derivative of \( y \) with respect to \( t \) is found by multiplying the derivative of \( y \) with respect to \( u \) by the derivative of \( u \) with respect to \( t \). Here's the formula:

• \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \)

This rule is handy when dealing with logarithms, as seen with the function \( y = \ln(7t^5) \). By setting \( u = 7t^5 \), you first differentiate \( y \) with respect to \( u \) (using the derivative of the logarithmic function) and then multiply it by the derivative of \( u \) with respect to \( t \). The chain rule makes it manageable to differentiate complex nested functions.

Logarithmic Differentiation

Logarithmic differentiation is a technique sometimes used to simplify derivatives, especially when dealing with complicated functions. It is particularly useful for functions that are products, quotients, or powers where both the base and the exponent are variables.
By taking the natural logarithm of both sides of an equation, you can leverage the properties of logarithms to transform the differentiation problem into one that is easier to solve.
Here's how logarithmic differentiation can be applied:

• Reduce the function using properties like \( \ln(a \cdot b) = \ln(a) + \ln(b) \) or \( \ln(a/b) = \ln(a) - \ln(b) \).
• Differentiate each part separately.
• Combine the derivatives back together to find the derivative of the original function.

For example, consider the function \( y = \ln(x \cdot (1-x)^8) \) which can be split into simpler terms using logarithmic properties. Then, differentiate each part separately before combining the results back together. This approach makes complex differentiation more intuitive and manageable.

Product Rule

The product rule is used to differentiate functions that are products of two or more functions. This rule is essential when functions are multiplied together, and you need to find their derivatives.
The formula for the product rule is simple. Given two functions \( f(x) \) and \( g(x) \), the derivative of their product is:

• \( \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \)

In essence, you differentiate the first function and multiply it by the second, then add it to the first function multiplied by the derivative of the second.
A practical example would be applying it to the function \( y = 5x^4 \ln(x^2) \) which can be thought of as a product of \( f(x) = 5x^4 \) and \( g(x) = \ln(x^2) \). By using the product rule, you differentiate each function and construct a full derivative expression. This allows for the study of more complex behaviors of functions and their rates of change.