136/MR00. problem / Math - Solved Math Problems

MR-136 / 136. problem

Determine the first derivative of the following function:

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

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Source: MathReference: Generic Math Textbook - 136.)

Mathematical analysis → Derivation → Derivative of exponential functions

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

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

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

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

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

$u^{'} = 2 e^{2 x} - e^{- 3 x} \cdot (- 3)$

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

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

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

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

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

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

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

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

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