Question

Differentiate ${\left(x\right)}^{tanx}+(tanx{)}^x$ w.r.t $x$

Answer 1

$y=x^{\tan x}+{\left(\tan x\right)}^x$

Let$u=x^{\tan x}$ and $v={\left(\tan x\right)}^x$

Now,$u=x^{\tan x}$

Taking $\log$ both sides, we get

$\log u=\tan x\log x$

Differentiating above equation w.r.t. $x$, we have

$\frac{1}{u}\frac{du}{dx}=\tan x\left(\frac{1}{x}\right)+\log x\left({\sec }^{2}x\right)$

$\frac{du}{dx}=u(\frac{\tan x}{x}+{\sec }^{2}x\log x)$

$\frac{du}{dx}=x^{\tan x}(\frac{\tan x}{x}+{\sec }^{2}x\log x)$

Similarly,

$v={\left(\tan x\right)}^x$

Taking $\log$ both sides, we have

$\log v=x\log \left(\tan x\right)$

$\frac{1}{v}\frac{dv}{dx}=\log \left(\tan x\right)\cdot 1+x\left(\frac{1}{\tan x}{\sec }^{2}x\right)$

$\frac{dv}{dx}=v(\log \left(\tan x\right)+\frac{x{\sec }^{2}x}{\tan x})$

$\frac{dv}{dx}={\left(\tan x\right)}^x(\log \left(\tan x\right)+\frac{x{\sec }^{2}x}{\tan x})$

Therefore,

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$\frac{dy}{dx}=x^{\tan x}(\frac{\tan x}{x}+{\sec }^{2}x\log x)+{\left(\tan x\right)}^x(\log \left(\tan x\right)+\frac{x{\sec }^{2}x}{\tan x})$