The given function is: y=(tanx)x. y = (\tan x)^x. y=(tanx)x. Take the natural logarithm on both sides to simplify the power: lny=xln(tanx). \ln y = x \ln (\tan x). lny=xln(tanx). Differentiate both sides with respect to xxx: 1y⋅dydx=ln(tanx)+x⋅1tanx⋅sec2x. \frac{1}{y} \cdot \frac{dy}{dx} = \ln (\tan x) + x \cdot \frac{1}{\tan x} \cdot \sec^2 x. y1⋅dxdy=ln(tanx)+x⋅tanx1⋅sec2x. Simplify: 1y⋅dydx=ln(tanx)+x⋅sec2xtanx. \frac{1}{y} \cdot \frac{dy}{dx} = \ln (\tan x) + x \cdot \frac{\sec^2 x}{\tan x}. y1⋅dxdy=ln(tanx)+x⋅tanxsec2x. Multiply through by y=(tanx)xy = (\tan x)^xy=(tanx)x: dydx=(tanx)x[ln(tanx)+x⋅sec2xtanx]. \frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \frac{\sec^2 x}{\tan x}\right]. dxdy=(tanx)x[ln(tanx)+x⋅tanxsec2x]. Simplify further: dydx=(tanx)x[ln(tanx)+x⋅1sinxcosx]. \frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \frac{1}{\sin x \cos x}\right]. dxdy=(tanx)x[ln(tanx)+x⋅sinxcosx1]. Final answer: dydx=(tanx)x[ln(tanx)+x⋅cscxsecx]. \frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \csc x \sec x\right]. dxdy=(tanx)x[ln(tanx)+x⋅cscxsecx].