Wasn't sure if i took the differentiation right!
heres the pic:
https://imgur.com/a/xubrz
Wasn't sure if i took the differentiation right!
heres the pic:
https://imgur.com/a/xubrz
No.
$\displaystyle \begin{align*} \dfrac{ d }{ dx } \left( x^{ \tan x } \right)^{ \sec x } & = \dfrac{d}{dx} \left( e^{ \ln x^{ \tan x \sec x } } \right) \\ & = \dfrac{d}{dx} \left( e^{ \tan x \sec x \ln x } \right) \\ & = e^{ \tan x \sec x \ln x } \dfrac{d}{dx} \left( \tan x \sec x \ln x \right) \\ & = x^{ \tan x \sec x } \left( \dfrac{ \sec x \tan x }{ x } + ( \sec^3 x + \sec x \tan^2 x ) \ln x \right) \end{align*}$
You had $\displaystyle \sec^2 x+\sec x \tan^2 x$, but it should be $\displaystyle \sec^3 x + \sec x \tan^2 x$
preference, imho ...
$y = x^{\tan{x} \cdot \sec{x}}$
$\ln{y} = \tan{x} \cdot \sec{x} \cdot \ln{x}$
$\dfrac{y'}{y} = \tan{x} \cdot \sec{x} \cdot \dfrac{1}{x} + \tan{x} \cdot \sec{x} \cdot \tan{x} \cdot \ln{x} + \sec^2{x} \cdot \sec{x} \cdot \ln{x}$
$\dfrac{y'}{y} = \sec{x} \bigg[\dfrac{\tan{x}}{x} + \ln{x} \left(2\tan^2{x}+1\right)\bigg]$
$y' = x^{\tan{x} \cdot \sec{x}} \cdot \sec{x} \bigg[\dfrac{\tan{x}}{x} + \ln{x} \left(2\tan^2{x}+1\right)\bigg]$
Because $\left(x^{\tan(x)}\right)^{\sec(x)}=\exp\left(\tan (x)\sec(x)\log(x)\right)$ we have
$\large\frac{d}{{dx}}{\left( {{x^{\tan (x)}}} \right)^{\sec (x)}} = \exp \left( {\tan (x)\sec (x)\log (x)} \right)\cdot\frac{d}{{dx}}\left( {\tan (x)\sec (x)\log (x)} \right)$
This is my preference.