Question

consider the following.

$f(x)=4x^{4}-9x^{3}$

find the following derivatives.

$f(x)=
$

$f(x)=
$

$f(x)=
$

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Explanation:

Step1: Apply power - rule for first - derivative

The power - rule states that if $y = ax^n$, then $y'=nax^{n - 1}$. For $f(x)=4x^{4}-9x^{3}$, we have $f'(x)=4\times4x^{4 - 1}-9\times3x^{3 - 1}=16x^{3}-27x^{2}$.

Step2: Apply power - rule for second - derivative

Differentiate $f'(x)=16x^{3}-27x^{2}$ using the power - rule. $f''(x)=16\times3x^{3 - 1}-27\times2x^{2 - 1}=48x^{2}-54x$.

Step3: Apply power - rule for third - derivative

Differentiate $f''(x)=48x^{2}-54x$ using the power - rule. $f'''(x)=48\times2x^{2 - 1}-54\times1x^{1 - 1}=96x - 54$.

Answer:

$f'(x)=16x^{3}-27x^{2}$
$f''(x)=48x^{2}-54x$
$f'''(x)=96x - 54$