Question

find the first and second derivatives.

$y = \frac{8x^{3}}{3}-5x + 7e^{x}$

$\frac{dy}{dx}=8x^{2}-5 + 7e^{x}$

$\frac{d^{2}y}{dx^{2}}=\square$

Explanation:

Step1: Differentiate the first derivative

Differentiate each term of \(\frac{dy}{dx}=8x^{2}-5 + 7e^{x}\) with respect to \(x\).
For the term \(8x^{2}\), use the power rule \(\frac{d}{dx}(ax^{n})=nax^{n - 1}\), where \(a = 8\) and \(n=2\). So \(\frac{d}{dx}(8x^{2})=2\times8x^{2-1}=16x\).
For the term \(-5\), since \(\frac{d}{dx}(c)=0\) (where \(c\) is a constant), \(\frac{d}{dx}(-5)=0\).
For the term \(7e^{x}\), use the rule \(\frac{d}{dx}(ae^{x})=ae^{x}\), so \(\frac{d}{dx}(7e^{x})=7e^{x}\).

Step2: Combine the results

Combine the derivatives of each term: \(\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(8x^{2})-\frac{d}{dx}(5)+\frac{d}{dx}(7e^{x})\)
\(\frac{d^{2}y}{dx^{2}}=16x+0 + 7e^{x}\)

Answer:

\(16x + 7e^{x}\)