Question

find the third derivative of the given function.
f(x) = 9x^5 - 7x^4 + 2x^2 - 3x + 1
f(x) =

Explanation:

Step1: Recall power - rule for derivatives

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$.

Step2: Find the first derivative $f^\prime(x)$

For $f(x)=9x^5 - 7x^4+2x^2 - 3x + 1$, using the power - rule:
$f^\prime(x)=\frac{d}{dx}(9x^5)-\frac{d}{dx}(7x^4)+\frac{d}{dx}(2x^2)-\frac{d}{dx}(3x)+\frac{d}{dx}(1)$
$=9\times5x^{5 - 1}-7\times4x^{4 - 1}+2\times2x^{2 - 1}-3\times1x^{1 - 1}+0$
$=45x^4-28x^3 + 4x-3$.

Step3: Find the second derivative $f^{\prime\prime}(x)$

Differentiate $f^\prime(x)=45x^4-28x^3 + 4x-3$:
$f^{\prime\prime}(x)=\frac{d}{dx}(45x^4)-\frac{d}{dx}(28x^3)+\frac{d}{dx}(4x)-\frac{d}{dx}(3)$
$=45\times4x^{4 - 1}-28\times3x^{3 - 1}+4\times1x^{1 - 1}-0$
$=180x^3-84x^2 + 4$.

Step4: Find the third derivative $f^{\prime\prime\prime}(x)$

Differentiate $f^{\prime\prime}(x)=180x^3-84x^2 + 4$:
$f^{\prime\prime\prime}(x)=\frac{d}{dx}(180x^3)-\frac{d}{dx}(84x^2)+\frac{d}{dx}(4)$
$=180\times3x^{3 - 1}-84\times2x^{2 - 1}+0$
$=540x^2-168x$.

Answer:

$540x^2-168x$