Question

Find the derivatives of the following functions. (a) \(f(x)=x^{2} e^{x} \cot x\) (b) \(g(x)=2^{x} 3^{x} 4^{x}\)

Short answer

(a) \(f'(x) = 2xe^x \cot x + x^2e^x \cot x - x^2e^x \csc^2 x\); (b) \(g'(x) = (24)^x \ln 24\)."

Step-by-step solution

Differentiate the first function

Given the function \(f(x)=x^{2} e^{x} \cot x\). We apply the product rule, which states that the derivative of a product \(u(x)v(x)w(x)\) is \(u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x)\). Here \(u(x) = x^2\), \(v(x) = e^x\), and \(w(x) = \cot x\).

Step 1.1: Differentiate each component of the product

Find the derivatives: - \(u'(x) = \frac{d}{dx}(x^2) = 2x\)- \(v'(x) = \frac{d}{dx}(e^x) = e^x\)- \(w'(x) = \frac{d}{dx}(\cot x) = -\csc^2 x\)

Step 1.2: Apply the product rule

Substitute the derivatives into the product rule: \(f'(x) = (2x)e^x \cot x + x^2(e^x \cot x) + x^2e^x(-\csc^2 x)\).Simplify to obtain:\[f'(x) = 2xe^x \cot x + x^2e^x \cot x - x^2e^x \csc^2 x.\]

Differentiate the second function

Given the function \(g(x) = 2^x 3^x 4^x\). Notice it can be rewritten as \(g(x) = (2\cdot3\cdot4)^x\). We use the property that the derivative of an exponential function \(a^x\) is \(a^x \ln a\).

Step 2.1: Simplify the function

Simplify \(g(x) = (24)^x\) and apply the exponential derivative rule.\(g'(x) = (24)^x \ln 24\).

Finalize the solution

Compile the derivatives obtained:- \(f'(x) = 2xe^x \cot x + x^2e^x \cot x - x^2e^x \csc^2 x\)- \(g'(x) = (24)^x \ln 24\).

Key concepts

Product Rule

One of the most essential techniques in calculus derivatives is the product rule. This rule is indispensable when you are faced with differentiating a function that is the product of multiple other functions. Imagine you are trying to differentiate a function composed of three separate parts, let's call them \( u(x) \), \( v(x) \), and \( w(x) \). The product rule helps by telling you exactly how to take the derivative of such a product efficiently.

The product rule is defined as follows for three functions:

• \((uvw)' = u'vw + uv'w + uvw'\)

This means that you take the derivative of each function one at a time while keeping the other two functions intact. Then, you sum up all the results. For example, if you need to differentiate \( f(x)=x^2 e^x \cot x \), you first identify each component:

• \( u(x) = x^2 \)
• \( v(x) = e^x \)
• \( w(x) = \cot x \)

After determining the individual derivatives of these functions:

• \( u'(x) = 2x \)
• \( v'(x) = e^x \)
• \( w'(x) = -\csc^2 x \)

You substitute these into the product rule to find the derivative. This eases up breaking down the complex product of functions into manageable parts.

Exponential Functions

Exponential functions exhibit one of the most straightforward yet powerful forms in calculus. They appear in various forms, but commonly, they look like \( a^x \) where \( a \) is a constant. The beauty of exponential functions lies in their unique differentiation property.

When differentiating an exponential function like \( a^x \), you use a simple formula:

• \( \frac{d}{dx}(a^x) = a^x \ln a \)

This means that the derivative of an exponential function is the function itself multiplied by the natural logarithm of the base. For example, if you have \( g(x) = 2^x 3^x 4^x \), you can simplify it to \( g(x) = (24)^x \) because \( 2 \cdot 3 \cdot 4 = 24 \).

By applying the derivative rule for exponential functions, the derivative becomes:

• \( g'(x) = (24)^x \ln 24 \)

Understanding this rule helps tackle problems involving growth rates, decay, and other phenomena easily described by exponential behaviors.

Trigonometric Derivatives

When it comes to trigonometric functions, their derivatives play a crucial role in integration and solving calculus problems. For our case, consider the function \( \cot x \), which is one of the trigonometric functions that may not be as commonly encountered as sine or cosine.

The derivative of \( \cot x \) is important to remember:

• \( \frac{d}{dx}(\cot x) = -\csc^2 x \)

This derivative tells us how the slope of the cotangent function changes at any given point on its curve. Differentiating trigonometric functions like \( \cot x \) is crucial in applying rules such as the product rule correctly.

When you're tasked to differentiate a complex product such as \( x^2 e^x \cot x \), knowing the derivative of each factor, including trigonometric ones, ensures that you apply these differentiation rules efficiently. Getting comfortable with these derivatives simplifies the entire differentiation process in calculus.