Let's perform the differentiation: <math><mo>(</mo><mi>x</mi><mo>·</mo><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup><mo>)'</mo></math>
For <math><mi>U1</mi><mo>·</mo><mi>V2</mi></math> = <math><mi>x</mi><mo>·</mo><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup></math>, use the product/quotient rule: <math><mo>(</mo><mi>U1</mi><mo>·</mo><mi>V2</mi><mo>)'</mo><mo> = </mo><mo>(</mo><mi>U1</mi><mo>)'</mo><mo>·</mo><mi>V2</mi><mo> + </mo><mi>U1</mi><mo>·</mo><mo>(</mo><mi>V2</mi><mo>)'</mo></math>
Let's perform the differentiation <math><mi>U1</mi></math>: <math><mo>(</mo><mi>x</mi><mo>)'</mo></math>
 <math><mn>1</mn></math>
Let's differentiate <math><mi>V2</mi></math>: <math><mo>(</mo><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup><mo>)'</mo></math>
 <math><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup></math>
Let's evaluate the resulting derivative (<math><mi>U1</mi><mo>·</mo><mi>V2</mi></math>)': <math><mn>1</mn><mo>·</mo><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup><mo> + </mo><mi>x</mi><mo>·</mo><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup></math>
 <math><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup><mo> + </mo><mi>x</mi><mo>·</mo><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup></math>
 <math><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup><mo>·</mo><mi>x</mi><mo> + </mo><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup></math>
This expression can be factorized into: <math><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup><mo>·</mo><mo>(</mo><mi>x</mi><mo> + </mo><mn>1</mn><mo>)</mo></math>

▷<b>Result: &nbsp; <math><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup><mo>·</mo><mo>(</mo><mi>x</mi><mo> + </mo><mn>1</mn><mo>)</mo></math></b><span style='font-size:smaller'> &nbsp; &nbsp; (computation required 10 steps and 7 ms)</span>