Question

if y = x tanx + secx , then find the value of dy/dx at x=pi/4

Answer 1

Dear student
$$\begin{gathered} \mathrm{Given}: \\ \mathrm{y}=\mathrm{xtanx}+\mathrm{secx} \\ \mathrm{Differentiate}\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x}\mathrm{we}\mathrm{get} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\mathrm{tanx}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{secx}\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{xsec}}^2\mathrm{x}+\mathrm{tanx}+\mathrm{secx}\mathrm{tanx} \\ {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{\mathrm{x}=\frac{\mathrm{\pi }}{4}}=\frac{\mathrm{\pi }}{4}{\sec }^2\left(\frac{\mathrm{\pi }}{4}\right)+\tan \left(\frac{\mathrm{\pi }}{4}\right)+\sec \left(\frac{\mathrm{\pi }}{4}\right)\tan \left(\frac{\mathrm{\pi }}{4}\right) \\ =\frac{\mathrm{\pi }}{4}{\left(\sqrt{2}\right)}^2+1+1\times \sqrt{2} \\ =\frac{2\mathrm{\pi }}{4}+1+\sqrt{2} \\ =\frac{\mathrm{\pi }}{2}+1+\sqrt{2} \\ =\frac{\mathrm{\pi }+2+2\sqrt{2}}{2} \end{gathered}$$
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