Possible derivation:

\(\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({e}^{2}{x}-{5}{y}\right)}\)

Differentiate the sum term by term and factor out constants:

\(\displaystyle={e}^{2}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}\right)}\right)}+\frac{d}{{\left.{d}{x}\right.}}{\left(-{5}{y}\right)}\)

The derivative of x is 1:

\(\displaystyle=\frac{d}{{\left.{d}{x}\right.}}{\left(-{5}{y}\right)}+{1}{e}^{2}\)

The derivative of -5 y is zero:

\(\displaystyle={e}^{2}+{0}\)

Simplify the expression:

Answer:

\(\displaystyle={e}^{2}\)