Problem 40 - Calculus - AP Edition

Question

Find \(d y / d x\) using any method. $$y=\left(x^{3}+\sqrt[3]{x}\right) 5^{x}$$

Step-by-Step Solution

Answer

\(d y / d x = (3x^2 + \frac{1}{3}x^{-2/3}) 5^x + (x^3 + \sqrt[3]{x}) 5^x \ln(5)\).

Step 1: Identify the Form

The given function \(y = (x^3 + \sqrt[3]{x}) 5^{x}\) is a product of two functions: \(u(x) = x^3 + \sqrt[3]{x}\) and \(v(x) = 5^x\). To find \(d y / d x\), we will use the product rule for derivatives.

Step 2: Differentiate \(u(x)\)

Differentiate \(u(x) = x^3 + \sqrt[3]{x}\). \(\frac{d}{dx}(x^3) = 3x^2\) and \(\frac{d}{dx}(x^{1/3}) = \frac{1}{3}x^{-2/3}\). Thus, \(u'(x) = 3x^2 + \frac{1}{3}x^{-2/3}\).

Step 3: Differentiate \(v(x)\)

Differentiate \(v(x) = 5^x\) using the exponential rule. The derivative of \(5^x\) is \(5^x \ln(5)\). Thus, \(v'(x) = 5^x \ln(5)\).

Step 4: Apply the Product Rule

The product rule states \((uv)' = u'v + uv'\).Substitute the derivatives found in Steps 2 and 3:\(d y / d x = (3x^2 + \frac{1}{3}x^{-2/3}) 5^x + (x^3 + \sqrt[3]{x}) 5^x \ln(5)\).

Step 5: Simplify the Expression

Combine the terms to simplify:\(d y / d x = 3x^2 \cdot 5^x + \frac{1}{3} x^{-2/3} \cdot 5^x + x^3 \cdot 5^x \ln(5) + x^{1/3} \cdot 5^x \ln(5)\).

Key Concepts

Product Rule

The product rule is used when you have to find the derivative of a product of two functions. It allows us to differentiate expressions like the one given in the exercise, where two separate functions are multiplied together, such as \(u(x)\) and \(v(x)\).

To apply the product rule, remember the formula:

• \((uv)' = u'v + uv'\)

It means you first find the derivative of the first function, multiplying it by the original second function. Then, add the product of the second function's derivative and the original first function.

In our exercise, \(u(x) = x^3 + \sqrt[3]{x}\) and \(v(x) = 5^x\). We calculate their derivatives separately, then use the product rule to find the overall derivative.

Exponential Rule

Exponential functions, like \(a^x\), have their own simple differentiation rule. When differentiating such functions, you multiply them by the natural logarithm (ln) of their base.

For \(a^x\), the derivative is calculated as:

• \(\frac{d}{dx}(a^x) = a^x \ln(a)\)

This rule is easy to apply once you identify the base and maintain consistency with geometrical growth rates.

In the exercise, we identified \(v(x) = 5^x\). Using the exponential rule, it's derivative is \(v'(x) = 5^x \ln(5)\). Using this, we can easily find how \(v(x)\) changes when differentiated.

Differentiation Techniques

Differentiating can seem tricky, but with techniques like the product rule and the exponential rule, you can tackle complex functions with ease. Combining simple differentiation rules lets you deal with more complex derivatives, even those involving products or composite functions.

Here’s how you can improve your differentiation skills:

• Break down the function into recognizable patterns or components that match known differentiation rules.
• Use a step-by-step approach to tackle different parts of the function before summarizing them by combining results.
• Practice regularly to recognize which techniques to apply in various scenarios easily.

By employing these methods, solving derivative problems like the given exercise becomes straightforward, and you begin developing an analytical approach within calculus. Insights on which rule to use ensure understanding and pave the way for tackling advanced differentiation problems confidently!