Differentiation of secx(tan√x)? In few steps.

Mar 24, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \left[\frac{{\sec}^{2} \sqrt{x}}{2 \sqrt{x}} + \tan \sqrt{x} \tan x\right]$

Explanation:

y=secx(tan√x)
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \frac{d}{\mathrm{dx}} \left(\tan \sqrt{x}\right) + \tan \sqrt{x} \frac{d}{\mathrm{dx}} \left(\sec x\right)$
$= \sec x {\sec}^{2} \sqrt{x} \frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) + \tan \sqrt{x} \sec x \tan x$
$= \sec x {\sec}^{2} \sqrt{x} \left(\frac{1}{2 \sqrt{x}}\right) + \tan \sqrt{x} \sec x \tan x$
$= \frac{1}{2 \sqrt{x}} \sec x {\sec}^{2} \sqrt{x} + \tan \sqrt{x} \sec x \tan x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \left[\frac{{\sec}^{2} \sqrt{x}}{2 \sqrt{x}} + \tan \sqrt{x} \tan x\right]$

Mar 24, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec x {\sec}^{2} \sqrt{x} \frac{\sqrt{x}}{2 x} + \tan \sqrt{x} \sec x \tan x$

Explanation:

.

$y = \sec x \tan \sqrt{x}$

let $\sqrt{x} = u = {x}^{\frac{1}{2}} , \therefore x = {u}^{2}$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}} = \frac{\sqrt{x}}{2 x}$

$d \left(\tan \sqrt{x}\right) = d \left(\tan u\right) = {\sec}^{2} u \mathrm{du}$

Let's divide both sides by$\mathrm{dx}$:

$\frac{d \left(\tan \sqrt{x}\right)}{\mathrm{dx}} = {\sec}^{2} u \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) = {\sec}^{2} \sqrt{x} \frac{\sqrt{x}}{2 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \left(\frac{d \left(\tan \sqrt{x}\right)}{\mathrm{dx}}\right) + \tan \sqrt{x} \sec x \tan x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec x {\sec}^{2} \sqrt{x} \frac{\sqrt{x}}{2 x} + \tan \sqrt{x} \sec x \tan x$