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Find the indicated derivative. \[ \frac{d y}{d x} \text { for } y=\frac{e^{x}}{2 x^{2}-13} \] \( \frac{d y}{d x}=\square \)

Ask by Garrett Sherman. in the United States
Mar 08,2025

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\[ \frac{dy}{dx} = \frac{e^x(2x^2 - 4x - 13)}{(2x^2 - 13)^2} \]

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To find the derivative \( \frac{d y}{d x} \) where \( y = \frac{e^{x}}{2 x^{2} - 13} \), you can use the quotient rule. The quotient rule states that if you have a function \( y = \frac{u}{v} \), the derivative \( y' \) is given by: \[ \frac{d y}{d x} = \frac{u'v - uv'}{v^2} \] Here, let \( u = e^{x} \) and \( v = 2x^{2} - 13 \). Now, calculate \( u' \) and \( v' \): 1. \( u' = e^x \) 2. \( v' = 4x \) Now, applying the quotient rule: \[ \frac{d y}{d x} = \frac{(e^{x})(2x^{2}-13) - (e^{x})(4x)}{(2x^{2}-13)^{2}} \] This simplifies to: \[ \frac{d y}{d x} = \frac{e^{x}((2x^{2}-13) - 4x)}{(2x^{2}-13)^{2}} = \frac{e^{x}(2x^{2} - 4x - 13)}{(2x^{2}-13)^{2}} \] Thus, the derivative is: \[ \frac{d y}{d x} = \frac{e^{x}(2x^{2}-4x-13)}{(2x^{2}-13)^{2}} \]

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