Question

if y=x^tanx +(tanx)^x, then find dy/dx​

Answer 1

$y = x {}^{ tanx } + tanx {}^{x}$

$y = u+ v\: \: \: \:$

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \: \: \: \: \: \: \: \: - (1)$

$u = x {}^{tanx}$

$log \: u = tanx \: logx$

Apply differentiation

$\frac{1}{u} \frac{du}{dx} = tanx( \frac{1}{x} ) + logx(sec {}^{2} x)$

$\frac{du}{dx} = x {}^{tanx}( \frac{tanx}{x} + sec {}^{2} x \: logx)$

Now,

$v = tanx {}^{x}$

Take log and apply diff.

$logv = xlog \: tanx$

$\frac{1}{v} \frac{dv}{dx} = 1(logtanx) + x( \frac{sec {}^{2}x }{tanx} )$

$\frac{dv}{dx} = tanx {}^{x}( logtanx + x\frac{sec {}^{2}x }{tanx} )$

Put in equation ( 1 )

$\frac{dy}{dx} = x {}^{tanx}( \frac{tanx}{x} + sec {}^{2} x \: logx) + tanx {}^{x}( logtanx + x\frac{sec {}^{2}x }{tanx} )$