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In this instance, both our reward as well as our divisor are portions. For that reason, we need to initially use the portion guidelines prior to acquiring them making use of the quotient formula:

\$\$ f( x) = frac {frac {x} {ln {(x)}}} {frac {x} {e ^ x}} = frac {xe ^ x} {x ln {(x)}} \$\$

Then, with this comes our quotient formula for features with numerous departments:

\$\$ F'( x) = frac {uv cdot (sw’+ ws’) hspace {2.3 pt}– hspace {2.3 pt} sw cdot (uv’+ vu’)} {(uv) ^ 2} \$\$

Step 1: Given that we have numerous departments in our provided feature, we will certainly stand for the numerator of the reward as \$latex s\$, the of the reward as \$latex u\$, the numerator of the divisor as \$latex v\$ as well as the of the divisor as \$latex w\$.

Therefore, we have

\$\$ f( x) = frac {frac {s} {u}} {frac {v} {w}} = frac {frac {x} {ln {(x)}}} {frac {x} {e ^ x}} \$\$

Applying the guidelines of portions:

\$\$ f( x) = frac {sw} {uv} = frac {xe ^ x} {x ln {(x)}} \$\$

Then, we have

\$ latex s = x\$
\$ latex u = ln {(x)} \$
\$ latex v = x\$
\$ latex w = e ^ x\$

Step 2: Set apart \$latex s\$, \$latex u\$, \$latex v\$ as well as \$latex w\$.

Hence, we have

\$ latex s = x\$
\$ latex s’ = 1\$

\$ latex u = ln {(x)} \$
\$ latex u’ = frac {1} {x} \$

\$ latex v = x\$
\$ latex v’ = 1\$

\$ latex w = e ^ x\$
\$ latex w’ = e ^ x\$

Step 3: Replacement \$latex s\$, \$latex u\$, \$latex v\$ as well as \$latex w\$ in the formula for the ratio guideline for features with numerous departments:

\$\$ F'( x) = frac {uv cdot (sw’+ ws’) hspace {2.3 pt}– hspace {2.3 pt} sw cdot (uv’+ vu’)} {(uv) ^ 2} \$\$

\$\$ frac {d} {dx} f( x) = frac {left[(x ln{(x)}) cdot ((x) cdot (e^x) + (e^x) cdot (1)) right] hspace {2.3 pt}– hspace {2.3 pt} left[(xe^x) cdot left((ln{(x)}) cdot (1) + (x) cdot left(frac{1}{x} right) right) right]} {(x ln {(x)} )^ 2} \$\$

Step 4: Simplify algebraically:

\$\$ frac {d} {dx} f( x) = frac {[(x ln{(x)}) cdot (xe^x + e^x)] hspace {2.3 pt}– hspace {2.3 pt} [(xe^x) cdot (ln{(x)} + 1)]} {(x ln {(x)} )^ 2} \$\$

\$\$ frac {d} {dx} f( x) = frac {(x ^ 2 e ^ x ln {(x)} + xe ^ x ln {(x)}) hspace {2.3 pt}– hspace {2.3 pt} (xe ^ x ln {(x)} + xe ^ x)} {(x ln {(x)} )^ 2} \$\$

\$\$ frac {d} {dx} f( x) = frac {x ^ 2 e ^ x ln {(x)}} {(x ln {(x)} )^ 2} + frac {xe ^ x ln {(x)}} {(x ln {(x)} )^ 2}– left( frac {xe ^ x ln {(x)}} {(x ln {(x)} )^ 2} + frac {xe ^ x} {(x ln {(x)} )^ 2} right)\$\$

\$\$ frac {d} {dx} f( x) = frac {e ^ x} {ln {(x)}} + frac {e ^ x} {x ln {(x)}}– frac {e ^ x} {x ln {(x)}}– frac {e ^ x} {x( ln {(x)} )^ 2} \$\$

\$\$ frac {d} {dx} f( x) = frac {e ^ x} {ln {(x)}}– frac {e ^ x} {x( ln {(x)} )^ 2} \$\$

\$\$ frac {d} {dx} f( x) = frac {xe ^ x ln {(x)}} {x( ln {(x)} )^ 2}– frac {e ^ x} {x( ln {(x)} )^ 2} \$\$

The last response is:

\$\$ f'( x) = frac {xe ^ x ln {(x)}– e ^ x} {x( ln {(x)} )^ 2} \$\$