Question

find the given higher - order derivative.
f(x)=x^{3}-x^{3}/4,f^{(3)}(x)
f^{(3)}(x)=

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1. 6 / 6 points

in the figure below, the graphs of f, f, and f are shown on the same set of coo

Explanation:

Step1: Recall derivative rule

The derivative of $x^n$ is $nx^{n - 1}$.

Step2: Differentiate $f'(x)=x^{3}-\frac{x^{3}}{4}=\frac{3x^{3}}{4}$ to get $f''(x)$

Using the power - rule, if $y = \frac{3x^{3}}{4}$, then $y'=f''(x)=\frac{3}{4}\times3x^{2}=\frac{9x^{2}}{4}$.

Step3: Differentiate $f''(x)$ to get $f^{(3)}(x)$

Differentiating $f''(x)=\frac{9x^{2}}{4}$ using the power - rule ($y = ax^{n}$, $y'=anx^{n - 1}$), we have $f^{(3)}(x)=\frac{9}{4}\times2x=\frac{9x}{2}$.

Answer:

$\frac{9x}{2}$