Question

derivative of 1-logx/1+logx w.r.t x​

Answer 1

$\textbf{Given:}$

$\mathsf{\dfrac{1-logx}{1+logx}}$

$\textbf{To find:}$

$\textsf{Derivative of}\;\mathsf{\dfrac{1-logx}{1+logx}}$

$\textbf{Solution:}$

$\mathsf{Consider,}$

$\mathsf{y=\dfrac{1-logx}{1+logx}}$

$\textsf{Using quotient rule,}$

$\boxed{\mathsf{\dfrac{d(\frac{u}{v})}{dx}=\dfrac{v\,\frac{du}{dx}-u\,\frac{dv}{dx}}{v^2}}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(1+logx)(\frac{-1}{x})-(1-logx)(\frac{1}{x})}{(1+logx)^2)}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{\frac{1}{x}[-(1+logx)-(1-logx)]}{(1+logx)^2)}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{\frac{1}{x}[-1-logx-1+logx]}{(1+logx)^2)}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{\frac{1}{x}(-2)}{(1+logx)^2)}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{\frac{-2}{x}}{(1+logx)^2)}}$

$\implies\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{-2}{x(1+logx)^2)}}}$

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