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Differentiate. \[ y=\left(9 x^{2}+8\right)(5 x+4) \] \( y^{\prime}=\square \)

Ask by Burns Bowman. in the United States
Mar 15,2025

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\( y' = 135x^2 + 72x + 40 \)

Solución

We are given \[ y = \left(9x^2+8\right)(5x+4). \] To differentiate, we use the product rule which states that \[ y' = u'v + uv', \] where \[ u = 9x^2+8 \quad \text{and} \quad v = 5x+4. \] 1. Differentiate \( u \): \[ u' = \frac{d}{dx}(9x^2+8)=18x. \] 2. Differentiate \( v \): \[ v' = \frac{d}{dx}(5x+4)=5. \] 3. Apply the product rule: \[ y' = (18x)(5x+4) + (9x^2+8)(5). \] 4. Expand and simplify: \[ (18x)(5x+4) = 90x^2 + 72x, \] and \[ (9x^2+8)(5) = 45x^2 + 40. \] Therefore, \[ y' = 90x^2 + 72x + 45x^2 + 40 = 135x^2 + 72x + 40. \] Thus, the derivative is \[ y' = 135x^2 + 72x + 40. \]

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To differentiate the function \( y = (9x^2 + 8)(5x + 4) \), we can apply the product rule, which states that if you have two functions \( u \) and \( v \), then \( (uv)' = u'v + uv' \). Let: - \( u = 9x^2 + 8 \) - \( v = 5x + 4 \) Now, differentiate each part: - \( u' = 18x \) - \( v' = 5 \) Using the product rule: \[ y' = u'v + uv' = (18x)(5x + 4) + (9x^2 + 8)(5) \] Now, substituting in known values: \[ y' = 18x(5x + 4) + (9x^2 + 8)(5) \] Distributing gives: \[ = 90x^2 + 72x + 45x^2 + 40 \] Combining like terms: \[ y' = 135x^2 + 72x + 40 \] Therefore, the final result is: \[ y' = 135x^2 + 72x + 40 \]

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