Exercise
$\frac{d}{dx}y=\frac{d}{dx}tanx^{\frac{1}{x}}$
Step-by-step Solution
Final answer to the exercise
$\frac{x^2\left(\left(2x\sec\left(x\right)^2\tan\left(x\right)-\ln\left(\tan\left(x\right)\right)\sec\left(x\right)^2\right)\tan\left(x\right)^{\left(\frac{1}{x}-1\right)}+\left(x\sec\left(x\right)^2-\ln\left(\tan\left(x\right)\right)\tan\left(x\right)\right)\left(\frac{-\ln\left(\tan\left(x\right)\right)}{x^2}+\left(\frac{1}{x}-1\right)\sec\left(x\right)\csc\left(x\right)\right)\tan\left(x\right)^{\left(\frac{1}{x}-1\right)}\right)+2x\left(-x\sec\left(x\right)^2+\ln\left(\tan\left(x\right)\right)\tan\left(x\right)\right)\tan\left(x\right)^{\left(\frac{1}{x}-1\right)}}{x^{4}}$