Question

find f(x), f(x), and f^{(3)}(x) for the following function. f(x)=8x^{3}+5x^{2}+6x. f(x)=

Explanation:

Step1: Apply power - rule for differentiation

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $f(x)=8x^{3}+5x^{2}+6x$, we differentiate each term separately.
For the first term $8x^{3}$, using the power - rule: $\frac{d}{dx}(8x^{3})=8\times3x^{3 - 1}=24x^{2}$.
For the second term $5x^{2}$, $\frac{d}{dx}(5x^{2})=5\times2x^{2 - 1}=10x$.
For the third term $6x$, $\frac{d}{dx}(6x)=6\times1x^{1 - 1}=6$.
So, $f^\prime(x)=24x^{2}+10x + 6$.

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

Differentiate each term of $f^\prime(x)=24x^{2}+10x + 6$ using the power - rule.
For the first term $24x^{2}$, $\frac{d}{dx}(24x^{2})=24\times2x^{2 - 1}=48x$.
For the second term $10x$, $\frac{d}{dx}(10x)=10\times1x^{1 - 1}=10$.
For the third term $6$ (a constant), $\frac{d}{dx}(6)=0$.
So, $f^{\prime\prime}(x)=48x + 10$.

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

Differentiate each term of $f^{\prime\prime}(x)=48x + 10$.
For the first term $48x$, $\frac{d}{dx}(48x)=48\times1x^{1 - 1}=48$.
For the second term $10$ (a constant), $\frac{d}{dx}(10)=0$.
So, $f^{(3)}(x)=48$.

Answer:

$f^\prime(x)=24x^{2}+10x + 6$, $f^{\prime\prime}(x)=48x + 10$, $f^{(3)}(x)=48$