Question

find dy/dx of y=x^tanx + √x+1/x​

Answer 1

$y = {x}^{tanx} + \sqrt{x + \frac{1}{x} } \\ let \: \: \: u = {x}^{tanx} \: \: \: \: v = \sqrt{x + \frac{1}{x} } \\ so \: \: \: \: y = u + v \\ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} ...(1) \\ \\ now \: \: \: \: u = {x}^{tanx} \\ applying \: log \: both \: \: sides \\ logu = tanxlogx \\ diff. \: \: \: w.r.t \: \: \: x \\ \frac{1}{u} \frac{du}{dx} = tanx \frac{1}{x} + logx( {sec}^{2} x) \\ \frac{du}{dx} = {x}^{tanx} ( \frac{tanx}{x} + {sec}^{2} x \: logx) \\ \\ now \: \: \: v = \sqrt{x + \frac{1}{x} } \\ \frac{dv}{dx} = \frac{1}{2 \sqrt{x + \frac{1}{x} } } (1 - \frac{1}{ {x}^{2} } ) \\ \\ from \: \: \: (1) \\ \frac{dy}{dx} = {x}^{tanx} ( \frac{tanx}{x} + {sec}^{2} x \: logx) + \frac{(1 - \frac{1}{ {x}^{2} }) }{2 \sqrt{x + \frac{1}{x} } }$