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  • If <img alt="y=\log \left(\frac{1+x}{1-x}\right)^{1 / 4}-\frac{1}{2} \cot ^{-1} x" src="https://entrancecorner.oncodecogs.com/gif.latex?y%3D%5Clog%20%5Cleft%28%5Cfrac%7B1&plus;x%7D%7B1-x%7

If y=\log \left(\frac{1+x}{1-x}\right)^{1 / 4}-\frac{1}{2} \cot ^{-1} x,then  \frac{d y}{d x}=

Option: 1

\frac{1}{1-x^4}


Option: 2

\frac{1}{1+x^4}


Option: 3

\frac{2}{1-x^4}


Option: 4

\frac{2}{1+x^4}


Answers (1)

best_answer

\\y=\log \left(\frac{1+x}{1-x}\right)^{1 / 4}-\frac{1}{2} \cot ^{-1} x \\y = \frac{1}{4}\log \left(\frac{1+x}{1-x}\right)-\frac{1}{2} \cot ^{-1} x \\y = \frac{1}{4}\log \left(1+x\right)-\frac{1}{4}\log \left(1-x\right)-\frac{1}{2} \cot ^{-1} x \\ \therefore \frac{dy}{dx}= \frac{1}{4(1+x)}+\frac{1}{4(1-x)}-\frac{1}{2} \cdot \frac{-1}{1+x^{2}} \\= \frac{(1-x)+(1+x)}{4(1-x^2)}+\frac{1}{2} \cdot \frac{1}{1+x^{2}} \\=\frac{1}{2(1-x^2)}+\frac{1}{2} \cdot \frac{1}{1+x^{2}} \\=\frac{1}{1-x^4}

Posted by

Devendra Khairwa

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