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Find the derivative of \( f(x)=x^{3} 7^{3 x} \) \( f^{\prime}(x)=\square \)

Ask by Love Powers. in the United States
Mar 12,2025

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The derivative of \( f(x) = x^{3} \cdot 7^{3x} \) is \( f'(x) = 3 \cdot 7^{3x} \left( x^{2} + x^{3} \ln(7) \right) \).

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To find the derivative of \( f(x) = x^{3} 7^{3x} \), we can apply the product rule. Let \( u = x^3 \) and \( v = 7^{3x} \). The product rule states that \( (uv)' = u'v + uv' \). 1. First, we find \( u' = 3x^2 \). 2. Next, we calculate \( v \). We can rewrite \( v \) using logarithmic differentiation: - \( v' = 7^{3x} \cdot \ln(7) \cdot 3 \). - Therefore, \( v' = 3 \ln(7) \cdot 7^{3x} \). Now, substituting these back into the product rule: \[ f'(x) = u'v + uv' = 3x^2 \cdot 7^{3x} + x^3 \cdot (3 \ln(7) \cdot 7^{3x}). \] Factoring out \( 7^{3x} \): \[ f'(x) = 7^{3x} \left( 3x^2 + 3x^3 \ln(7) \right). \] Thus, the final answer is: \[ f'(x) = 7^{3x} \left( 3x^2 + 3x^3 \ln(7) \right). \]

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