Sendo \(f(x,y) = { e^{2x} \over y}\), a expressão de \({df \over dx}\) é:

\(\Longrightarrow {df \over dx} = {d \over dx} \Big ( { e^{2x} \over y} \Big )\)

\(\Longrightarrow {df \over dx} = {1 \over y} {d \over dx}( e^{2x} )\)

\(\Longrightarrow {df \over dx} = {1 \over y} \Big ( e^{2x}\cdot {d (2x)\over dx} \Big )\)

\(\Longrightarrow {df \over dx} = {1 \over y} ( e^{2x}\cdot 2 )\)

\(\Longrightarrow \fbox {$ {df \over dx} = {2 e^{2x} \over y} $}\)

Sendo \(f(x,y) = { e^{2x} \over y}\), a expressão de \({df \over dy}\) é:

\(\Longrightarrow {df \over dy} = {d \over dy} \Big ( { e^{2x} \over y} \Big )\)

\(\Longrightarrow {df \over dy} = e^{2x}{d \over dy} \Big ( y^{-1} \Big )\)

\(\Longrightarrow {df \over dy} = e^{2x}( -1 \cdot y^{-1-1})\)

\(\Longrightarrow {df \over dy} = e^{2x}( - y^{-2})\)

\(\Longrightarrow \fbox { $ {df \over dy} = -{e^{2x} \over y^2} $}\)

∂f/∂x = 2e^(2x)/y

∂f/∂y = -(e^x/y)^2