Question

find the derivative of the following function f(x) w.r.t x at x=1
cos^-1[sin route over of 1+x/2]+x^x

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{y}={\cos }^{-1}\left(\sin \sqrt{1+\frac{\mathrm{x}}{2}}\right)+{\mathrm{x}}^{\mathrm{X}} \\ \mathrm{Let}\mathrm{u}={\cos }^{-1}\left(\sin \sqrt{1+\frac{\mathrm{x}}{2}}\right)\mathrm{and}\mathrm{v}={\mathrm{x}}^{\mathrm{X}} \\ \frac{\mathrm{du}}{\mathrm{dx}}=\frac{-1}{\sqrt{1-{\left(\sin \sqrt{1+\frac{\mathrm{x}}{2}}\right)}^2}}\times \frac{\mathrm{d}}{\mathrm{dx}}\sin \sqrt{1+\frac{\mathrm{x}}{2}} \\ =\frac{-1}{\sqrt{1-\left({\sin }^2\sqrt{1+\frac{\mathrm{x}}{2}}\right)}}\times \cos \sqrt{1+\frac{\mathrm{x}}{2}}\times \frac{1}{4\sqrt{1+\frac{\mathrm{x}}{2}}} \\ =\frac{-1}{\cos \sqrt{1+\frac{\mathrm{x}}{2}}}\times \cos \sqrt{1+\frac{\mathrm{x}}{2}}\times \frac{1}{4\sqrt{1+\frac{\mathrm{x}}{2}}} \\ =\frac{-1}{4\sqrt{1+\frac{\mathrm{x}}{2}}} \\ \mathrm{v}={\mathrm{x}}^{\mathrm{x}} \\ \mathrm{logv}=\mathrm{xlogx} \\ \mathrm{differentiating},\mathrm{we}\mathrm{get} \\ \frac{1}{\mathrm{v}}\frac{\mathrm{dv}}{\mathrm{dx}}=1+\mathrm{logx} \\ \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}\left(1+\mathrm{logx}\right)={\mathrm{x}}^{\mathrm{x}}\left(1+\mathrm{logx}\right) \\ \mathrm{Now}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-1}{4\sqrt{1+\frac{\mathrm{x}}{2}}}+{\mathrm{x}}^{\mathrm{x}}\left(1+\mathrm{logx}\right) \\ \mathrm{Now}\mathrm{at}\mathrm{x}=1 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{4\sqrt{1+\frac{1}{2}}}+1^1\left(1+\log 1\right)=\frac{-\sqrt{2}}{4\sqrt{3}}+1=1-\frac{\sqrt{6}}{12} \end{gathered}$$