find the derivative of the function.\ny = x^3 - \\frac{x^2}{28}+8x + 4\ny=\\square

Explanation:

Step1: Apply power - rule to $x^3$

The power - rule states that if $y = x^n$, then $y'=nx^{n - 1}$. For $y_1=x^3$, $y_1'=3x^{3-1}=3x^2$.

Step2: Apply power - rule to $-\frac{x^2}{28}$

For $y_2 =-\frac{x^2}{28}=-\frac{1}{28}x^2$, using the power - rule, $y_2'=-\frac{1}{28}\times2x^{2 - 1}=-\frac{1}{14}x$.

Step3: Apply power - rule to $8x$

For $y_3 = 8x=8x^1$, by the power - rule, $y_3'=8\times1x^{1 - 1}=8$.

Step4: Derivative of a constant

The derivative of a constant $y_4 = 4$ is $y_4' = 0$.

Step5: Sum of derivatives

Since $y=y_1 + y_2+y_3 + y_4$, then $y'=y_1'+y_2'+y_3'+y_4'$. $y'=3x^2-\frac{1}{14}x + 8+0$.

Answer:

$3x^2-\frac{1}{14}x + 8$