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find the derivatives of Y with respect to X ,

y=ln1/x(x+1)

Apr 30, 2020

#1
+30637
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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
+30637
+3

"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