Question

find the first and second derivatives.
y = \frac{6x^{5}}{5}-3x + 8e^{x}
\frac{dy}{dx}=\square

Explanation:

Step1: Differentiate each term

Use power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$ and $\frac{d}{dx}(e^x)=e^x$. For $\frac{6x^5}{5}$, $n = 5$ and $a=\frac{6}{5}$, so its derivative is $5\times\frac{6}{5}x^{5 - 1}=6x^4$. For $-3x$, $n = 1$ and $a=-3$, so its derivative is $-3$. For $8e^x$, its derivative is $8e^x$.

Step2: Combine the derivatives of each term

$\frac{dy}{dx}=6x^4-3 + 8e^x$

Step3: Differentiate the first - derivative to get the second - derivative

Differentiate $6x^4-3 + 8e^x$ term - by - term. For $6x^4$, using the power - rule again, $n = 4$ and $a = 6$, so its derivative is $4\times6x^{4 - 1}=24x^3$. The derivative of a constant $-3$ is $0$. The derivative of $8e^x$ is $8e^x$.

Step4: Combine the derivatives of the terms of the first - derivative

$\frac{d^2y}{dx^2}=24x^3+8e^x$

Answer:

First derivative: $\frac{dy}{dx}=6x^4-3 + 8e^x$
Second derivative: $\frac{d^2y}{dx^2}=24x^3+8e^x$