Odrediti prvi izvod sledeće funkcije:

$y=\frac{{e}^{2x}-{e}^{-3x}}{{e}^{5x}-{e}^{7x}}$

$y=\frac{{e}^{2x}-{e}^{-3x}}{{e}^{5x}-{e}^{7x}}$

$y=\frac{{{e}}^{2x}{-}{{e}}^{-3x}}{{{e}}^{5x}{-}{{e}}^{7x}}=\frac{{u}}{{v}}$

${u}{=}{{e}}^{2x}{-}{{e}}^{-3x}$

${v}{=}{{e}}^{5x}{-}{{e}}^{7x}$

${y}^{\text{'}}=\frac{{u}^{\text{'}}·v-u·{v}^{\text{'}}}{{v}^{2}}$

${u}^{\text{'}}=2{e}^{2x}-{e}^{-3x}·\left(-3\right)$

${u}^{\text{'}}=2{e}^{2x}+3{e}^{-3x}$

${v}^{\text{'}}=5{e}^{5x}-7{e}^{7x}$

${y}^{\text{'}}=\frac{\left(2{e}^{2x}+3{e}^{-3x}\right)\left({e}^{5x}-{e}^{7x}\right)-\left({e}^{2x}-{e}^{-3x}\right)\left(5{e}^{5x}-7{e}^{7x}\right)}{{\left({e}^{5x}-{e}^{7x}\right)}^{2}}$

${y}^{\text{'}}=\frac{{2}{{e}}^{7x}{-}{2}{{e}}^{9x}{+}{3}{{e}}^{2x}{-}{3}{{e}}^{4x}{-}{5}{{e}}^{7x}{+}{7}{{e}}^{9x}{+}{5}{{e}}^{2x}{-}{7}{{e}}^{4x}}{{\left({e}^{5x}\right)}^{2}-2{e}^{5x}{e}^{7x}+{\left(-{e}^{7x}\right)}^{2}}$

${y}^{\text{'}}=\frac{{5}{{e}}^{9x}{-}{3}{{e}}^{7x}{-}{10}{{e}}^{4x}{+}{8}{{e}}^{2x}}{{e}^{10x}-2{e}^{12x}+{e}^{14x}}$

${y}^{\text{'}}=\frac{{{e}}^{2x}\left(5{e}^{7x}-3{e}^{5x}-10{e}^{2x}+8\right)}{{{e}}^{10x}\left(1-2{e}^{2x}+{e}^{4x}\right)}$

${y}^{\text{'}}=\frac{5{e}^{7x}-3{e}^{5x}-10{e}^{2x}+8}{{e}^{8x}\left(1-2{e}^{2x}+{e}^{4x}\right)}$

${y}^{\text{'}}=\frac{5{e}^{7x}-3{e}^{5x}-10{e}^{2x}+8}{{e}^{8x}{\left(1-{e}^{2x}\right)}^{2}}$