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# Ask AI: find the derivative of y=(e^-2x)((1-e^-2x)^-1/2) ...with e being eulers constant

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Question: find the derivative of y=(e^-2x)((1-e^-2x)^-1/2) ...with e being eulers constant

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To find the derivative of the function $$y = (e^{-2x}) \left(1 - e^{-2x}\right)^{-\frac{1}{2}}$$, we need to use the product rule as well as the chain rule. Let's denote the function as $$y = u \cdot v$$, where $u = e^{-2x}$ $v = \left(1 - e^{-2x}\right)^{-\frac{1}{2}}$ First, we need to find the derivatives of $$u$$ and $$v$$. ### 1. Derivative of $$u = e^{-2x}$$ Using the chain rule for the exponential function: $\frac{du}{dx} = \frac{d}{dx}\left(e^{-2x}\right) = e^{-2x} \cdot (-2) = -2e^{-2x}$ ### 2. Derivative of $$v = \left(1 - e^{-2x}\right)^{-\frac{1}{2}}$$ Let's use the chain rule for this part also. Let $$w = 1 - e^{-2x}$$, so $$v = w^{-\frac{1}{2}}$$. First, find $$\frac{dw}{dx}$$: $\frac{dw}{dx} = \frac{d}{dx} \left(1 - e^{-2x}\right) = 0 - (-2)e^{-2x} = 2e^{-2x}$ Now, find $$\frac{dv}{dw}$$: $\frac{dv}{dw} = \frac{d}{dw}\left(w^{-\frac{1}{2}}\right) = -\frac{1}{2} w^{-\frac{3}{2}}$ Now use the chain rule to find $$\frac{dv}{dx}$$: $\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx} = -\frac{1}{2} \left(1 - e^{-2x}\right)^{-\frac{3}{2}} \cdot 2e^{-2x} = -\left(1 - e^{-2x}\right)^{-\frac{3}{2}} e^{-2x}$ ### 3. Apply the Product Rule The product rule states that if $$y = u \cdot v$$, then $\frac{dy}{dx} = u'v + uv'$ We already found $$u' = -2e^{-2x}$$ and $$v' = -\left(1 - e^{-2x}\right)^{-\frac{3}{2}} e^{-2x}$$. Therefore, $\frac{dy}{dx} = (u'v) + (uv')$ $\frac{dy}{dx} = (-2e^{-2x}) \left(1 - e^{-2x}\right)^{-\frac{1}{2}} + (e^{-2x}) \left[-\left(1 - e^{-2x}\right)^{-\frac{3}{2}} e^{-2x} \right]$ Simplify this expression: $\frac{dy}{dx} = -2e^{-2x}\left(1 - e^{-2x}\right)^{-\frac{1}{2}} - e^{-4x}\left(1 - e^{-2x}\right)^{-\frac{3}{2}}$ Thus, the derivative of the given function is: $\frac{dy}{dx} = -2e^{-2x}\left(1 - e^{-2x}\right)^{-\frac{1}{2}} - e^{-4x}\left(1 - e^{-2x}\right)^{-\frac{3}{2}}$

May 14, 2024, 5:09 p.m.

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This article was generated in part by one of OpenAI’s large-scale language-generation model, answering user paul_anthony's question. Images were generated with OpenAI's AI model, DALL·E 2. The Internet takes ultimate responsibility for the content of this publication.