Derivative of x^5(3-6x^-9)?

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Tony B Share
Feb 9, 2018

${f}^{'} \left(x\right) = 15 {x}^{4} + 24 {x}^{- 5}$

Explanation:

Set $y = {x}^{5} \left(3 - 6 {x}^{- 9}\right)$

Multiply out the brackets

$y = 3 {x}^{5} - \frac{6 {x}^{5}}{x} ^ 9$

$y = 3 {x}^{5} - \frac{6}{x} ^ 4$

$y = 3 {x}^{5} - 6 {x}^{- 4}$

Or if you prefer

$f \left(x\right) = 3 {x}^{5} - 6 {x}^{- 4}$

The next step is to differentiate each term.

Using the principle that if: $g \left(x\right) = a {x}^{n} \to {g}^{'} \left(x\right) = \left(a \times n\right) {x}^{n - 1}$

${f}^{'} \left(x\right) = \left(3 \times 5\right) {x}^{5 - 1} - \left[6 \times \left(- 4\right) {x}^{- 4 - 1}\right]$

${f}^{'} \left(x\right) = 15 {x}^{4} + 24 {x}^{- 5}$

They used to use $\frac{\mathrm{dy}}{\mathrm{dx}}$ for ${f}^{'} \left(x\right)$ You will still see this sometimes.
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$\textcolor{b l u e}{\text{Check}}$
Output from Maple:

$5 {x}^{4} \left(3 - \frac{6}{x} ^ 9\right) + \frac{54}{x} ^ 5$

$15 {x}^{4} - \frac{30}{x} ^ 5 + \frac{54}{x} ^ 5$

$15 {x}^{4} - \frac{24}{x} ^ 5$

$15 {x}^{4} - 24 {x}^{- 5} \textcolor{red}{\leftarrow \text{ Confirmed}}$

Then teach the underlying concepts
Don't copy without citing sources
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Explanation

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1s2s2p Share
Feb 9, 2018

$f ' \left(x\right) = 54 {x}^{-} 5 + 5 {x}^{4} \left(3 - 6 {x}^{-} 9\right)$

Explanation:

We have $f \left(x\right) = {x}^{5} \left(3 - 6 {x}^{-} 9\right) = g \left(x\right) h \left(x\right)$

$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + g ' \left(x\right) h \left(x\right)$

$g \left(x\right) = {x}^{5}$
$g ' \left(x\right) = 5 {x}^{4}$

$h \left(x\right) = 3 - 6 {x}^{-} 9$
$h ' \left(x\right) = - 9 \cdot - 6 {x}^{- 9 - 1} = 54 {x}^{-} 10$

$f ' \left(x\right) = {x}^{5} \left(54 {x}^{-} 10\right) + 5 {x}^{4} \left(3 - 6 {x}^{-} 9\right)$
$\textcolor{w h i t e}{f ' \left(x\right)} = 54 {x}^{-} 5 + 5 {x}^{4} \left(3 - 6 {x}^{-} 9\right)$

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