Answer :

[tex]y=x^{\tan x}=e^{\ln x^{\tan x}}=e^{\tan x \ln x}\\ y'=e^{\tan x \ln x}\cdot(\sec^2 x\ln x+\tan x \cdot \frac{1}{x})\\ y'=x^{\tan x}(\sec^2 x\ln x+\frac{\tan x}{x})\\[/tex]
Answer is in attachment below. Please open it up in a new window to see it in full.
View image Аноним

Other Questions