Question

$$ \text { } , \text { find the indicated derivative. } $$ $$ D_{z}\left[3^{z} \ln (z+5)\right] $$

Short answer

The derivative is \( D_z[3^z \ln(z+5)] = 3^z \left( \ln(3) \ln(z+5) + \frac{1}{z+5} \right) \).

Step-by-step solution

Identify the need for the Product Rule

The function given is a product of two functions: \( f(z) = 3^z \) and \( g(z) = \ln(z+5) \). To find the derivative of the product, we need to use the Product Rule: \( (f \cdot g)' = f' \cdot g + f \cdot g' \).

Find the derivative of \( f(z) = 3^z \)

The derivative of an exponential function \( a^x \) is given by \( a^x \ln(a) \). Therefore, the derivative of \( f(z) = 3^z \) is \( f'(z) = 3^z \ln(3) \).

Find the derivative of \( g(z) = \ln(z+5) \)

The derivative of \( \ln(u) \) where \( u = z+5 \) is \( \frac{1}{u}\cdot \frac{du}{dz} \). Since \( u = z+5 \), \( \frac{du}{dz} = 1 \). Therefore, \( g'(z) = \frac{1}{z+5} \).

Apply the Product Rule

Using the Product Rule, \( (f \cdot g)' = f' \cdot g + f \cdot g' \), substitute \( f(z) = 3^z \), \( f'(z) = 3^z \ln(3) \), \( g(z) = \ln(z+5) \), and \( g'(z) = \frac{1}{z+5} \). This gives: \[ D_z[3^z \ln(z+5)] = 3^z \ln(3) \cdot \ln(z+5) + 3^z \cdot \frac{1}{z+5} \].

Simplify the expression

Factor out \( 3^z \) to obtain: \[ D_z[3^z \ln(z+5)] = 3^z \left( \ln(3) \cdot \ln(z+5) + \frac{1}{z+5} \right) \]. This is the final expression for the derivative.

Key concepts

Product Rule

Let's dive into the interesting world of derivatives with the Product Rule! When we have a function that is the product of two distinct functions, we use the Product Rule to find its derivative.

Imagine that you have two functions, say \( f(z) \) and \( g(z) \). The Product Rule states that the derivative of \( f(z) \cdot g(z) \) is found by this formula:

• \( (f \cdot g)' = f' \cdot g + f \cdot g' \)

This means, we take the derivative of the first function and multiply it by the second, then add the first function multiplied by the derivative of the second.

This allows us to methodically find the derivative of the entire expression—a process that becomes intuitive with practice. If you ever see a product of two functions and you need the derivative, the Product Rule is your go-to tool!

Exponential Functions

Exponential functions are a staple in calculus, and understanding how to differentiate them is crucial. Consider the function \( f(z) = 3^z \). This is an exponential function where the base is a constant (3) and the exponent is a variable \( z \).

To find the derivative of an exponential function like \( a^x \), we use the formula:

• \( (a^x)' = a^x \ln(a) \)

For our specific function \( 3^z \), applying this rule gives us the derivative \( f'(z) = 3^z \ln(3) \).

The logarithm of the base comes into play here (\( \ln(3) \) in our case), which is a unique characteristic of the derivative of exponential functions. Keep in mind, despite their complexity, exponential functions have a tidy pattern to their derivatives once you get the hang of the math.

Logarithmic Functions

Logarithmic functions, often involving expressions like \( \ln(z+5) \), have their own set of rules for differentiation.

The derivative of a logarithmic function is determined using the formula:

• For \( \ln(u) \), the derivative is \( \frac{1}{u} \cdot \frac{du}{dz} \)

In our earlier example, where \( g(z) = \ln(z+5) \), we set \( u = z+5 \), which means that \( \frac{du}{dz} = 1 \).

Therefore, the derivative is simply \( g'(z) = \frac{1}{z+5} \). This method is straightforward by isolating and translating the inside expression. Remember, the natural logarithm \( \ln(u) \) turns differentiation into a manageable one-step process once the expression inside is defined.

Differentiation Techniques

Differentiation techniques, such as the Product Rule and chain techniques for exponential and logarithmic functions, form the backbone of calculus. These techniques help solve a broad range of mathematical problems by breaking down complex expressions into manageable parts.

There's often an initial pause when you see a new problem, but by systematically applying these rules, differentiation becomes easier.

• The Product Rule helps when dealing with products of two separate functions.
• Derivatives of exponential functions incorporate the natural log of the base.
• Logarithmic derivatives simplify the inner function first for clarity.

With practice, these core techniques turn into skillful maneuvers that can tackle diverse calculus challenges with precision. Remember, the more you practice, the more intuitive these techniques become.