Question

17 multiple choice 1 point find the derivative. y = e^{x^4} ln x \bigcirc \frac{e^{x^4} + 4x^4 e^{x^4} ln x}{x} \bigcirc \frac{4x^4 e^{x^4} + 1}{x} \bigcirc \frac{e^{x^4} + 4x^3 e^{x^4} ln x}{x} \bigcirc \frac{e^{x^4} + 4e^{x^4} ln x}{x}

Explanation:

Step1: Identify the product rule

The function \( y = e^{x^4} \ln x \) is a product of two functions, \( u = e^{x^4} \) and \( v = \ln x \). The product rule states that \( (uv)' = u'v + uv' \).

Step2: Find \( u' \)

For \( u = e^{x^4} \), use the chain rule. Let \( t = x^4 \), so \( u = e^t \). Then \( u' = e^t \cdot t' = e^{x^4} \cdot 4x^3 \) (since \( t' = 4x^3 \)).

Step3: Find \( v' \)

For \( v = \ln x \), the derivative \( v' = \frac{1}{x} \).

Step4: Apply the product rule

\( y' = u'v + uv' = (4x^3 e^{x^4}) \ln x + e^{x^4} \cdot \frac{1}{x} \).

Step5: Combine terms over a common denominator

Factor out \( e^{x^4} \) and get a common denominator of \( x \):
\( y' = \frac{e^{x^4} + 4x^3 e^{x^4} \ln x}{x} \) (by multiplying the first term by \( \frac{x}{x} \) to combine: \( \frac{4x^3 e^{x^4} \ln x \cdot x}{x} + \frac{e^{x^4}}{x} = \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \)? Wait, no, wait. Wait, \( 4x^3 e^{x^4} \ln x \) times \( x \) over \( x \) is \( 4x^4 e^{x^4} \ln x \) over \( x \), and \( e^{x^4}/x \) is as is. Wait, no, my mistake in Step 4. Wait, \( u'v = 4x^3 e^{x^4} \ln x \), and \( uv' = e^{x^4} \cdot \frac{1}{x} \). So to combine, we get \( \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \)? Wait, no, \( 4x^3 e^{x^4} \ln x = \frac{4x^4 e^{x^4} \ln x}{x} \), and \( e^{x^4}/x \) is \( e^{x^4}/x \). So adding them gives \( \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \)? Wait, but looking at the options, the first option has \( 4x^4 e^{x^4} \ln x \), the third has \( 4x^3 \). Wait, I must have made a mistake. Wait, no, \( u = e^{x^4} \), \( u' = 4x^3 e^{x^4} \) (correct, chain rule: derivative of \( x^4 \) is \( 4x^3 \)). Then \( u'v = 4x^3 e^{x^4} \ln x \), \( uv' = e^{x^4} \cdot \frac{1}{x} \). Now, to combine, we can write \( 4x^3 e^{x^4} \ln x = \frac{4x^4 e^{x^4} \ln x}{x} \) (multiply numerator and denominator by \( x \)), and \( e^{x^4}/x \) is \( e^{x^4}/x \). So adding them: \( \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \), which is the first option? Wait, no, the first option is \( \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \), the third option is \( \frac{e^{x^4} + 4x^3 e^{x^4} \ln x}{x} \). Wait, I see my mistake. In Step 4, \( u'v = 4x^3 e^{x^4} \ln x \), and \( uv' = e^{x^4}/x \). So when combining, it's \( \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \)? Wait, no, \( 4x^3 e^{x^4} \ln x = \frac{4x^4 e^{x^4} \ln x}{x} \) (because \( 4x^3 \times x = 4x^4 \)), so yes, the first option is \( \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \). Wait, but let's recheck. Wait, the first option is \( \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \), the third is \( \frac{e^{x^4} + 4x^3 e^{x^4} \ln x}{x} \). Wait, my chain rule: \( u = e^{x^4} \), so \( u' = e^{x^4} \times 4x^3 \), correct. Then \( u'v = 4x^3 e^{x^4} \ln x \), and \( uv' = e^{x^4}/x \). So to combine, we have \( \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \) (since \( 4x^3 e^{x^4} \ln x = \frac{4x^4 e^{x^4} \ln x}{x} \)). So the first option is correct? Wait, no, the first option's numerator is \( e^{x^4} + 4x^4 e^{x^4} \ln x \), which matches. Wait, but in my initial calculation, I thought \( 4x^3 \times x = 4x^4 \), so yes. So the first option is correct. Wait, but let's check the options again. The first option is \( \frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x} \), the third is \( \frac{e^{x^4} + 4x^3 e^{x^4} \ln x}{x} \). Wait, I think I messed up the chain rule. Wait, \( u = e^{x^4} \), derivative is \( e^{x^4} \times 4x^3 \), correct. Then \( u'v = 4x^3 e^{x^4} \ln x \), and \( uv' = e^{x^4}/x \). So when we write \( 4x^3 e^{x^4} \ln x = \frac{4x^4 e^{x^4} \ln x}{x} \) (multiplying numerator and denominator by \( x \)), so the numerator becomes \( e^{x^4} + 4x^4 e^{x^4} \ln x \), so the first option is correct. Wait, but the third option has \( 4x^3 \) in the numerator. So maybe my mistake was in the common denominator. Wait, no, \( 4x^3 e^{x^4} \ln x + \frac{e^{x^4}}{x} = \frac{4x^4 e^{x^4} \ln x + e^{x^4}}{x} = \frac{e^{x^4}(1 + 4x^4 \ln x)}{x} \), which is the first option. So the first option is correct.

Answer:

The first option: \(\frac{e^{x^4} + 4x^4 e^{x^4} \ln x}{x}\) (the first circle option)