Question

find (\frac{d^2 y}{dx^2}). ( y = \frac{7x^5}{5} - 3x ) (\frac{d^2 y}{dx^2} = square)

Explanation:

Step1: Find the first derivative $\frac{dy}{dx}$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n - 1}$, for $y=\frac{7x^{5}}{5}-3x$:
The derivative of $\frac{7x^{5}}{5}$ is $\frac{7}{5}\times5x^{4}=7x^{4}$, and the derivative of $-3x$ is $- 3$. So $\frac{dy}{dx}=7x^{4}-3$.

Step2: Find the second derivative $\frac{d^{2}y}{dx^{2}}$

Differentiate $\frac{dy}{dx}=7x^{4}-3$ with respect to $x$. Using the power rule again, the derivative of $7x^{4}$ is $7\times4x^{3} = 28x^{3}$, and the derivative of $-3$ (a constant) is $0$. So $\frac{d^{2}y}{dx^{2}}=28x^{3}$.

Answer:

$28x^{3}$