Question

find the first and second derivatives.
y = \frac{8x^{3}}{3} - 5x + 7e^{x}
\frac{dy}{dx} = \square

Explanation:

Step1: Differentiate each term

Differentiate \(\frac{8x^{3}}{3}\), \(-5x\), and \(7e^{x}\) separately.
For \(\frac{8x^{3}}{3}\), use the power rule \((x^{n})^\prime = nx^{n - 1}\). The derivative is \(\frac{8\times3x^{2}}{3}=8x^{2}\).
For \(-5x\), using the power rule (\(x^{1}\) term), the derivative is \(-5\).
For \(7e^{x}\), since \((e^{x})^\prime=e^{x}\), the derivative is \(7e^{x}\).

Step2: Combine the derivatives

By the sum - rule of differentiation \((u + v+w)^\prime=u^\prime + v^\prime+w^\prime\), where \(u = \frac{8x^{3}}{3}\), \(v=-5x\), \(w = 7e^{x}\).
\(\frac{dy}{dx}=8x^{2}-5 + 7e^{x}\)

Answer:

\(8x^{2}-5 + 7e^{x}\)