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
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.