find f(x), f(x), and f^{(3)}(x) for the following function. f(x)=2x^{3}+8x^{2}+4x

Explanation:

Step1: Apply power - rule for differentiation

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $f(x)=2x^{3}+8x^{2}+4x$, $f^\prime(x)=\frac{d}{dx}(2x^{3})+\frac{d}{dx}(8x^{2})+\frac{d}{dx}(4x)$. $ \frac{d}{dx}(2x^{3})=2\times3x^{2}=6x^{2}$, $\frac{d}{dx}(8x^{2})=8\times2x = 16x$, $\frac{d}{dx}(4x)=4$. So, $f^\prime(x)=6x^{2}+16x + 4$.

Step2: Differentiate $f^\prime(x)$ to get $f^{\prime\prime}(x)$

Differentiate $f^\prime(x)=6x^{2}+16x + 4$ using the power - rule. $\frac{d}{dx}(6x^{2})=6\times2x=12x$, $\frac{d}{dx}(16x)=16$, $\frac{d}{dx}(4)=0$. So, $f^{\prime\prime}(x)=12x + 16$.

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

Differentiate $f^{\prime\prime}(x)=12x + 16$ using the power - rule. $\frac{d}{dx}(12x)=12$, $\frac{d}{dx}(16)=0$. So, $f^{(3)}(x)=12$.

Answer:

$f^\prime(x)=6x^{2}+16x + 4$, $f^{\prime\prime}(x)=12x + 16$, $f^{(3)}(x)=12$