if (f(x)=2x^{2}-x^{3}), find (f^{prime}(x), f^{primeprime}(x), f^{primeprimeprime}(x)), and…

if (f(x)=2x^{2}-x^{3}), find (f^{prime}(x), f^{primeprime}(x), f^{primeprimeprime}(x)), and (f^{(4)}(x)).\n\ngraph (f, f^{prime}, f^{primeprime}), and (f^{primeprimeprime}) on a common screen.\n\nare the graphs consistent with the geometric interpretations of these derivatives?\nthe graphs select consistent with the geometric interpretations of the derivatives because (f^{prime}) select where (f) has a slope
Answer
Explanation:
Step1: Apply power - rule for first derivative
The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $f(x)=2x^{2}-x^{3}$, we have $f^\prime(x)=\frac{d}{dx}(2x^{2})-\frac{d}{dx}(x^{3})$. Using the power - rule, $\frac{d}{dx}(2x^{2}) = 2\times2x^{2 - 1}=4x$ and $\frac{d}{dx}(x^{3})=3x^{2}$. So, $f^\prime(x)=4x - 3x^{2}$.
Step2: Apply power - rule for second derivative
Differentiate $f^\prime(x)=4x - 3x^{2}$ with respect to $x$. $\frac{d}{dx}(4x)=4$ and $\frac{d}{dx}(3x^{2}) = 6x$. So, $f^{\prime\prime}(x)=4 - 6x$.
Step3: Apply power - rule for third derivative
Differentiate $f^{\prime\prime}(x)=4 - 6x$ with respect to $x$. $\frac{d}{dx}(4)=0$ and $\frac{d}{dx}(6x)=6$. So, $f^{\prime\prime\prime}(x)=-6$.
Step4: Apply power - rule for fourth derivative
Differentiate $f^{\prime\prime\prime}(x)=-6$ with respect to $x$. Since $-6$ is a constant, $\frac{d}{dx}(-6)=0$. So, $f^{(4)}(x)=0$.
Answer:
$f^\prime(x)=4x - 3x^{2}$ $f^{\prime\prime}(x)=4 - 6x$ $f^{\prime\prime\prime}(x)=-6$ $f^{(4)}(x)=0$