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 term $8x^{3}$, using the power - rule: $(8x^{3})^\prime=8\times3x^{3 - 1}=24x^{2}$.
For the term $5x^{2}$, using the power - rule: $(5x^{2})^\prime=5\times2x^{2 - 1}=10x$.
For the term $6x$, using the power - rule: $(6x)^\prime=6\times1x^{1 - 1}=6$.
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 term $24x^{2}$, $(24x^{2})^\prime=24\times2x^{2 - 1}=48x$.
For the term $10x$, $(10x)^\prime=10\times1x^{1 - 1}=10$.
For the constant term $6$, $(6)^\prime = 0$.
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 term $48x$, $(48x)^\prime=48\times1x^{1 - 1}=48$.
For the constant term $10$, $(10)^\prime = 0$.
So, $f^{(3)}(x)=48$.

Answer:

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