Question

Find \(\frac{{{d^{\bf{9}}}}}{{d{x^{\bf{9}}}}}\left( {{x^{\bf{8}}}\,{\bf{ln}}\,x} \right)\).

Short answer

The value of \({D^9}y\) is \(\frac{{8!}}{x}\).

Step-by-step solution

Differentiate the function \(y = {x^{\bf{8}}}\,{\bf{ln}}\,x\)

The derivativeof the function \(y = {x^8}\ln x\) with respect to x can be calculated as follows:

\(\begin{aligned}y' &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{x^8}\ln x} \right)\\ &= \left( {\ln x} \right) \cdot \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{x^8}} \right) + \left( {{x^8}} \right) \cdot \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\ln x} \right)\\ &= 8{x^7}\left( {\ln x} \right) + {x^8}\left( {\frac{1}{x}} \right)\\ &= 8{x^7}\left( {\ln x} \right) + {x^7}\end{aligned}\)

Write the expression for \({D^{\bf{9}}}y\)

The expression for \({D^9}y\) can be expressed as:

\(\begin{aligned}{D^9}y &= {D^8}y'\\ &= {D^8}\left( {8{x^7}\ln x + {x^7}} \right)\end{aligned}\)

The 8^th derivative of \({x^7}\) is 0;

\({D^9}y = {D^8}\left( {8{x^7}\ln x} \right)\)

Further differentiate the equation to find \({D^{\bf{8}}}\)

Differentiate the equation \({D^9}y = {D^8}\left( {8{x^7}\ln x} \right)\).

\(\begin{aligned}{D^9}y &= {D^8}\left( {8{x^7}\ln x} \right)\\ &= {D^7}\left( {8 \cdot 7{x^6}\ln x + 8{x^6}} \right)\\ &= {D^7}\left( {8 \cdot 7{x^6}\ln x} \right)\\ &= {D^6}\left( {8 \cdot 7 \cdot 6{x^5}\ln x} \right)\\ &= D\left( {8!{x^0}\ln x} \right)\\ &= \frac{{8!}}{x}\end{aligned}\)

Thus, the value of \({D^9}y\) is \(\frac{{8!}}{x}\).