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Find the derivative (2) \( y=7 x^{3}-3 x^{2}+8 \)

Ask by Kelly Lee. in the United States
Mar 11,2025

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Answer

The derivative of \( y = 7x^3 - 3x^2 + 8 \) is \( y' = 21x^2 - 6x \).

Solution

1. Given the function: \[ y = 7x^3 - 3x^2 + 8 \] 2. Differentiate each term using the power rule \(\frac{d}{dx}(x^n) = nx^{n-1}\): - For \(7x^3\): \[ \frac{d}{dx}(7x^3) = 7 \cdot 3x^{3-1} = 21x^2 \] - For \(-3x^2\): \[ \frac{d}{dx}(-3x^2) = -3 \cdot 2x^{2-1} = -6x \] - For \(8\) (a constant): \[ \frac{d}{dx}(8) = 0 \] 3. Combine the results: \[ y' = 21x^2 - 6x \]

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To find the derivative of the function \( y = 7x^3 - 3x^2 + 8 \), we apply the power rule for differentiation. The derivative of \( ax^n \) is \( n \cdot ax^{n-1} \). So, differentiating \( y \) term by term, we get: - The derivative of \( 7x^3 \) is \( 21x^2 \), - The derivative of \( -3x^2 \) is \( -6x \), - The derivative of the constant \( 8 \) is \( 0 \). Combining these results, the derivative is: \( y' = 21x^2 - 6x \). Now you can use this to analyze the behavior of the function! The derivative helps you find the slopes of the tangent line at any point on the curve. If you're exploring local maxima or minima of the function, setting \( y' = 0 \) lets you locate critical points. These points are often where the graph of the function takes a beautiful turn—like a dancer changing direction mid-performance!

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