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  • Show that f(x) = 2x plus cot inverse x plus log(root 1 plus x^2) minus x is increasing in R.

Show that f(x)=2 x+\cot ^{-1} x+\log \left(\sqrt{1+x^{2}}-x\right) is increasing in R.

Answers (1)

Given: f(x)=2 x+\cot ^{-1} x+\log \left(\sqrt{1+x^{2}}-x\right)

To show: the mentioned function increases in R

Explanation: Given f(x)=2 x+\cot ^{-1} x+\log \left(\sqrt{1+x^{2}}-x\right)

Substituting the first derivative in respect to x

\\\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(2 \mathrm{x}+\cot ^{-1} \mathrm{x}+\log \left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)\right)}{\mathrm{dx}}$\\ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}\left(\cot ^{-1} \mathrm{x}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\log \left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)\right)}{\mathrm{dx}}
But the derivative of 2x is 2,so
\mathrm{f}^{\prime}(\mathrm{x})=2+\frac{\mathrm{d}\left(\cot ^{-1} \mathrm{x}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\log \left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)\right)}{\mathrm{dx}}

But the derivative of \cot ^{-1} \mathrm{x}=-\frac{1}{\mathrm{x}^{2}+1},so

\\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{\mathrm{d}\left(\log \left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)\right)}{\mathrm{dx}}$\\ $f^{\prime}(x)=2+\left(-\frac{1}{x^{2}+1}\right)+\frac{1}{\sqrt{1+x^{2}}-x} \cdot \frac{d\left(\sqrt{1+x^{2}}-x\right)}{d x}

When the sum rule is applied to the last part we get,

\\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{\mathrm{d}\left(\sqrt{1+\mathrm{x}^{2}}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}\right) \\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{\mathrm{d}\left(\left(1+\mathrm{x}^{2}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}-1\right) \\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{1}{2} \cdot\left(1+\mathrm{x}^{2}\right)^{\frac{1}{2}-1} \cdot \frac{\mathrm{d}\left(1+\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}-1\right) \\

\\\mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{1}{2} \cdot\left(1+\mathrm{x}^{2}\right)^{-\frac{1}{2}} \cdot\left[\frac{\mathrm{d}(1)}{\mathrm{d} \mathrm{x}}+\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}\right]-1\right) \\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{1}{2 \sqrt{1+\mathrm{x}^{2}}} \cdot[0+2 \mathrm{x}]-1\right) \\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{2 \mathrm{x}}{2 \sqrt{1+\mathrm{x}^{2}}}-1\right)

\\ \mathrm{f}^{\prime}(\mathrm{x})=2-\frac{1}{\mathrm{x}^{2}+1}+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}-1\right)$ $\\\\\mathrm{f}^{\prime}(\mathrm{x})=2-\frac{1}{\mathrm{x}^{2}+1}+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{\mathrm{x}-\sqrt{1+\mathrm{x}^{2}}}{\sqrt{1+\mathrm{x}^{2}}}\right)$ \\Cutting out all the like terms, $\mathrm{f}^{\prime}(\mathrm{x})=2-\frac{1}{\mathrm{x}^{2}+1}+\frac{-1}{\sqrt{1+\mathrm{x}^{2}}}$ \\Then, adding by taking the LCM, we get \\$\mathrm{f}^{\prime}(\mathrm{x})=\frac{2+2 \mathrm{x}^{2}-1-\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}^{2}+1}$\\ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}^{2}+1-\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}^{2}+1}

Then to calculate any real value x, the above value of f(x) is larger than or equal to zero

Hence \frac{2 x^{2}+1-\sqrt{1+x^{2}}}{x^{2}+1} \forall x \in R

And we know, if f'(x)\geq 0, then f(x) is increasing function.

Hence, the given function is an increasing function in R.

Posted by

infoexpert22

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