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find the derivatives of Y with respect to X ,

y=ln1/x(x+1)

 Apr 30, 2020

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 #1
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"find the derivatives of Y with respect to X , y=ln1/x(x+1)"

 

I assume this is \(y=\ln{\frac{1}{x(x+1)}}\)  which can be expressed as  \(y = -\ln{x}-\ln{(x+1)}\) using the properties of logarithms.

 

So we have:  \(\frac{dy}{dx} = -\frac{1}{x}-\frac{1}{x+1}\)  or  \(\frac{dy}{dx} = -\frac{2x+1}{x(x+1)}\)

 Apr 30, 2020
 #1
avatar+30637 
+3
Best Answer

"find the derivatives of Y with respect to X , y=ln1/x(x+1)"

 

I assume this is \(y=\ln{\frac{1}{x(x+1)}}\)  which can be expressed as  \(y = -\ln{x}-\ln{(x+1)}\) using the properties of logarithms.

 

So we have:  \(\frac{dy}{dx} = -\frac{1}{x}-\frac{1}{x+1}\)  or  \(\frac{dy}{dx} = -\frac{2x+1}{x(x+1)}\)

Alan Apr 30, 2020

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