Question

Evaluate dy/dx when y=cos^-1(log₂x)

Evaluate dy/dx when y = log[ √(1+x²)+ x / √(1+x²)-x]

Answer 1

$$\begin{gathered} 1) \\ \mathrm{y}={\cos }^{-1}\left({\log }_2\mathrm{x}\right) \\ \mathrm{So}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{\sqrt{1-{\left({\log }_2\mathrm{x}\right)}^2}}\times \frac{\mathrm{d}}{\mathrm{dx}}\left({\log }_2\mathrm{x}\right) \\ =\frac{-1}{\sqrt{1-{\left({\log }_2\mathrm{x}\right)}^2}}\times \frac{1}{\mathrm{xlog}2} \\ 2) \\ \mathrm{y}=\log \left[\frac{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}-\mathrm{x}}\times \frac{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\right] \\ =\log \left[\frac{{\left(\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}\right)}^2}{{\left(\sqrt{1+{\mathrm{x}}^2}\right)}^2-{\mathrm{x}}^2}\right]=\log \left[\frac{{\left(\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}\right)}^2}{1+{\mathrm{x}}^2-{\mathrm{x}}^2}\right]=\log \left[\frac{{\left(\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}\right)}^2}{1}\right] \\ =2\log \left(\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}\right) \\ \mathrm{So}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\times \frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}\right) \\ =\frac{2}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\times \left(\frac{\mathrm{d}}{\mathrm{dx}}\sqrt{1+{\mathrm{x}}^2}+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right) \\ =\frac{2}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\times \left(\frac{2\mathrm{x}}{2\sqrt{1+{\mathrm{x}}^2}}+1\right) \\ =\frac{2}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\times \frac{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}} \\ =\frac{2}{\sqrt{1+{\mathrm{x}}^2}} \end{gathered}$$