Answer

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Hint: Here, the given problem can be solved by simplifying the given function first

and then applying the suitable formulae of differentiation.

Given,

${x^{ - 4}}(3 - 4{x^{ - 5}}) \to (1)$

Let us simply the equation (1), we get

$3{x^{ - 4}} - 4{x^{ - 9}} \to (2)$

Now, we need to find the differentiation of equation (2) with respect to x i.e..,

$\begin{gathered}

\Rightarrow \frac{d}{{dx}}(3{x^{ - 4}} - 4{x^{ - 9}}) \\

\Rightarrow \frac{d}{{dx}}(3{x^{ - 4}}) - \frac{d}{{dx}}(4{x^{ - 9}}) \\

\Rightarrow 3\frac{d}{{dx}}({x^{ - 4}}) - 4\frac{d}{{dx}}({x^{ - 9}}) \\

\end{gathered} $

As we know that$\frac{d}{{dx}}({x^n}) = n.{x^{n - 1}}$.So applying the formulae, we get

$\begin{gathered}

\Rightarrow (3( - 4){x^{ - 4 - 1}}) - (4( - 9){x^{ - 9 - 1}}) \\

\Rightarrow - 12{x^{ - 5}} + 36{x^{ - 10}} \\

\end{gathered} $

Therefore$\frac{d}{{dx}}({x^{ - 4}}(3 - 4{x^{ - 5}})) = - 12{x^{ - 5}} + 36{x^{ - 10}}$.

Note: The differentiation formula of ${x^n}$i.e.., $\frac{d}{{dx}}({x^n}) = n.{x^{n - 1}}$can be

used for any value of n i.e.., it will be applicable even the value of n is positive, negative or

fractional value.

and then applying the suitable formulae of differentiation.

Given,

${x^{ - 4}}(3 - 4{x^{ - 5}}) \to (1)$

Let us simply the equation (1), we get

$3{x^{ - 4}} - 4{x^{ - 9}} \to (2)$

Now, we need to find the differentiation of equation (2) with respect to x i.e..,

$\begin{gathered}

\Rightarrow \frac{d}{{dx}}(3{x^{ - 4}} - 4{x^{ - 9}}) \\

\Rightarrow \frac{d}{{dx}}(3{x^{ - 4}}) - \frac{d}{{dx}}(4{x^{ - 9}}) \\

\Rightarrow 3\frac{d}{{dx}}({x^{ - 4}}) - 4\frac{d}{{dx}}({x^{ - 9}}) \\

\end{gathered} $

As we know that$\frac{d}{{dx}}({x^n}) = n.{x^{n - 1}}$.So applying the formulae, we get

$\begin{gathered}

\Rightarrow (3( - 4){x^{ - 4 - 1}}) - (4( - 9){x^{ - 9 - 1}}) \\

\Rightarrow - 12{x^{ - 5}} + 36{x^{ - 10}} \\

\end{gathered} $

Therefore$\frac{d}{{dx}}({x^{ - 4}}(3 - 4{x^{ - 5}})) = - 12{x^{ - 5}} + 36{x^{ - 10}}$.

Note: The differentiation formula of ${x^n}$i.e.., $\frac{d}{{dx}}({x^n}) = n.{x^{n - 1}}$can be

used for any value of n i.e.., it will be applicable even the value of n is positive, negative or

fractional value.

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