Question

Differentiate w.r.t $x$,

$y=$$tan\left\{\frac{tanx-1}{\sqrt{2tanx}}\right\}$

Answer 1

Differentiating with respect to $x$

$y=\tan \left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\tan \left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\right)={\sec }^2\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\frac{d}{dx}\left(\tan \left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\right)={\sec }^2\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\left(\frac{\sqrt{2\tan x}\frac{d}{dx}\left(\tan x-1\right)-\left(\tan x-1\right)\frac{d}{dx}\left(\sqrt{2\tan x}\right)}{{\left(\sqrt{2\tan x}\right)}^2}\right)={\sec }^2\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\left(\frac{\sqrt{2\tan x}\left({\sec }^{2}x\right)-\left(\tan x-1\right)\frac{1}{2\sqrt{2\tan x}}\frac{d}{dx}\left(2\tan x\right)}{\sqrt{2\tan x}}\right)={\sec }^2\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\left(\frac{\sqrt{2\tan x}\left({\sec }^{2}x\right)-\left(\tan x-1\right)\frac{2.{\sec }^{2}x}{2\sqrt{2\tan x}}}{\sqrt{2\tan x}}\right)={\sec }^2\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\left(\frac{2\tan x{\sec }^{2}x-\left(\tan x-1\right){\sec }^{2}x}{{\left(\sqrt{2\tan x}\right)}^{\frac{3}{2}}}\right)={\sec }^2\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\left(\frac{{\sec }^{2}x\left(2\tan x-\tan x+1\right)}{{\left(\sqrt{2\tan x}\right)}^{\frac{3}{2}}}\right)={\sec }^2\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)\left(\frac{{\sec }^{2}x(1+\tan x)}{{\left(\sqrt{2\tan x}\right)}^{\frac{3}{2}}}\right)$