Questions: Differentiate. y=x^5 ln x-frac13 x^3 fracd yd x=square

Differentiate. y=x^5 ln x-frac13 x^3 fracd yd x=square
Transcript text: Differentiate. \[ \begin{array}{l} y=x^{5} \ln x-\frac{1}{3} x^{3} \\ \frac{d y}{d x}=\square \end{array} \]
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Solution

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Solution Steps

To differentiate the given function \( y = x^5 \ln x - \frac{1}{3} x^3 \), we will apply the product rule and the power rule. The product rule is used for differentiating products of two functions, and the power rule is used for differentiating terms of the form \( x^n \).

  1. Identify the two parts of the function: \( u = x^5 \) and \( v = \ln x \).
  2. Apply the product rule: \((uv)' = u'v + uv'\).
  3. Differentiate each part: \( u' = 5x^4 \) and \( v' = \frac{1}{x} \).
  4. Differentiate the second term \(-\frac{1}{3} x^3\) using the power rule.
  5. Combine the results to find \(\frac{dy}{dx}\).
Step 1: Define the Function

We start with the function given by \[ y = x^5 \ln x - \frac{1}{3} x^3. \]

Step 2: Differentiate the First Term

Using the product rule for the first term \(x^5 \ln x\), we have: \[ \frac{d}{dx}(x^5 \ln x) = \frac{d}{dx}(u \cdot v) = u'v + uv', \] where \(u = x^5\) and \(v = \ln x\). Thus, \[ u' = 5x^4 \quad \text{and} \quad v' = \frac{1}{x}. \] This gives us: \[ \frac{d}{dx}(x^5 \ln x) = 5x^4 \ln x + x^4. \]

Step 3: Differentiate the Second Term

For the second term \(-\frac{1}{3} x^3\), we apply the power rule: \[ \frac{d}{dx}\left(-\frac{1}{3} x^3\right) = -\frac{1}{3} \cdot 3x^2 = -x^2. \]

Step 4: Combine the Results

Now, we combine the derivatives from Steps 2 and 3: \[ \frac{dy}{dx} = 5x^4 \ln x + x^4 - x^2. \]

Final Answer

Thus, the derivative of the function is \[ \boxed{\frac{dy}{dx} = 5x^4 \ln x + x^4 - x^2}. \]

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