Problem 39 - Calculus - AP Edition

Question

Find \(d y / d x\) using any method. $$y=\left(x^{2}+\sqrt{x}\right) 3^{x}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 3^x \left( 2x + \frac{1}{2}x^{-1/2} + (x^2 + \sqrt{x}) \ln(3) \right) \)

Step 1: Identify the Differentiation Rule

To find \( \frac{dy}{dx} \), notice that your function \( y = (x^2 + \sqrt{x}) 3^x \) is a product of two functions: \( u(x) = x^2 + \sqrt{x} \) and \( v(x) = 3^x \). Hence, we will use the product rule for differentiation, which states: \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \).

Step 2: Differentiate the First Function

Differentiate \( u(x) = x^2 + \sqrt{x} \). Use the power rule, \( \frac{d}{dx}[x^n] = nx^{n-1} \).- For \( x^2 \), \( u'(x) = 2x \).- For \( \sqrt{x} = x^{1/2} \), \( u'(x) = \frac{1}{2}x^{-1/2} \).So, \( u'(x) = 2x + \frac{1}{2}x^{-1/2} \).

Step 3: Differentiate the Second Function

Differentiate \( v(x) = 3^x \) using the exponential differentiation rule. The derivative of \( a^x \) where \( a \) is a constant is \( a^x \ln(a) \).So, \( v'(x) = 3^x \ln(3) \).

Step 4: Apply the Product Rule

Now, apply the product rule to find \( \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \).- \( u'(x)v(x) = \left(2x + \frac{1}{2}x^{-1/2}\right)3^x \)- \( u(x)v'(x) = (x^2 + \sqrt{x}) 3^x \ln(3) \)Combine these results:\[ \frac{dy}{dx} = \left(2x + \frac{1}{2}x^{-1/2}\right)3^x + (x^2 + \sqrt{x}) 3^x \ln(3) \]

Step 5: Simplify the Expression

Factor out \( 3^x \) from both terms:\[ \frac{dy}{dx} = 3^x \left( 2x + \frac{1}{2}x^{-1/2} + (x^2 + \sqrt{x}) \ln(3) \right) \]

Key Concepts

Differentiating Exponential Functions

Exponential functions play a crucial role in calculus, and understanding how to differentiate them is important. When you have a function in the form \( a^x \), where \( a \) is a constant, the derivative is given by the formula \( a^x \ln(a) \). This means that to differentiate an exponential function, you:

• Keep the base \( a^x \) the same as the original function.
• Multiply the whole expression by the natural logarithm of the base, \( \ln(a) \).

So, in the case of \( v(x) = 3^x \), the derivative is \( v'(x) = 3^x \ln(3) \). Recognizing this pattern is key to mastering problems that involve exponential functions.

Understanding the Power Rule

The power rule is one of the simplest, yet most powerful tools in differentiation. It states: if you have \( x^n \), then its derivative is \( nx^{n-1} \). This rule can be applied to various powers of \( x \), including fractions and negative powers.

• If \( u(x) = x^2 \), you apply the power rule to get \( u'(x) = 2x \).
• For \( u(x) = \sqrt{x} = x^{1/2} \), apply the power rule to yield \( u'(x) = \frac{1}{2}x^{-1/2} \). This shows how the rule works even with fractional exponents.

By breaking down each part of the function effectively, you can handle complex differentiation problems with ease.

Simplifying Derivatives

After applying differentiation rules, the resulting expression may look complicated. Simplifying derivatives is vital for both understanding and practical applications. Once you've applied the necessary differentiation rules to each part of the function, you:

• Look for common factors in each term of the derivative.
• Factor these common elements out to simplify the expression.

In the current problem, after applying the product rule, we combined terms and factored out \( 3^x \) since it’s a common factor. This resulted in the expression \[ \frac{dy}{dx} = 3^x \left( 2x + \frac{1}{2}x^{-1/2} + (x^2 + \sqrt{x}) \ln(3) \right) \]. Factoring helps in making the result easier to read and can make subsequent calculations more straightforward.