Math, asked by dee0, 1 year ago

ex+ey=ex+y prove that dy/dx=ex(ey-1)/ey(ex-1)​

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Answered by Anonymous
17

Answer \:  \\  \\ e {}^{x}  + e {}^{y}  = e {}^{(x + y)}  \\  \\ Differentiate \:  \: Both \: sides\: with \: \\  respect \: to \:  \: x \: we \: have \\  \\  \frac{d(e {}^{x} )}{dx}  +  \frac{d(e {}^{y} )}{dx}  =  \frac{d(e {}^{(x + y)} )}{dx}  \\  \\ e {}^{x}  + e {}^{y}  \times  \frac{dy}{dx}  = e {}^{(x + y)}  \times (1 +  \frac{dy}{dx} ) \\  \\ e {}^{x}  - e {}^{(x +y )}  = e {}^{(x + y)}  \:  \:  \:  \frac{dy}{dx}  - e {}^{y}  \:  \:  \:  \frac{dy}{dx}  \\  \\ e {}^{x}  - e {}^{(x + y)}  =  \frac{dy}{dx}  \:  \: (e {}^{(x + y)}   - e {}^{y} ) \\  \\  \frac{dy}{dx}  =  \frac{e {}^{x} - e {}^{(x + y)}  }{e {}^{(x + y)} - e {}^{y}  }  \\  \\  \frac{dy}{dx}  =  \frac{e {}^{x}  - e {}^{x}  \times e {}^{y} } {e {}^{x}  \times e {}^{y}  - e {}^{y} }  \\  \\  \frac{dy}{dx}  =   \frac{e {}^{x} (1 - e {}^{y} )}{e {}^{y}(e {}^{x}  - 1) }  \\  \\  \\ therefore \:  \: Differentiation \: of \:  \:  \\  \\ e {}^{x}  + e {}^{y}  = e {}^{(x + y)}  is \\  \\  \frac{e {}^{x} (1 - e {}^{y} )}{e {}^{y}(e {}^{x}   - 1)}

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