Question

can anyone show me how to solve this?​

Answer 1

$lny = x \times ln \: (tanx )\\ \\ \frac{dlny}{dy} \times \frac{dy}{dx} = ln(tanx) \times 1 + x \times \frac{1}{tanx} \times {sec}^{2} x \\ \\ \frac{1}{y} \times \frac{dy}{dx} = ln(tanx) + \frac{x. {sec}^{2} x}{tanx} \\ \\ \frac{dy}{dx } = y(ln \: tanx + \frac{x. {sec}^{2}x }{tanx} ) \\ \\ now \: put \: the \: \: value \: of \: y = {tanx}^{x} \\ \\ we \: get \\ \\ \frac{dy}{dx} = {tanx}^{x} (ln \: tanx + \frac{x. {sec}^{2}x }{tanx} )$