Math / CalculusQuestiondifferentiats cosx\cos ^{x}cosxStuddy SolutionSolve for dydx \frac{dy}{dx} dxdy by multiplying both sides by y y y:dydx=y(−xtan(x)+ln(cos(x))) \frac{dy}{dx} = y \left(-x \tan(x) + \ln(\cos(x))\right) dxdy=y(−xtan(x)+ln(cos(x)))Substitute back y=(cos(x))x y = (\cos(x))^x y=(cos(x))x:dydx=(cos(x))x(−xtan(x)+ln(cos(x))) \frac{dy}{dx} = (\cos(x))^x \left(-x \tan(x) + \ln(\cos(x))\right) dxdy=(cos(x))x(−xtan(x)+ln(cos(x))) The derivative of y=cosx y = \cos^x y=cosx is:dydx=(cos(x))x(−xtan(x)+ln(cos(x))) \frac{dy}{dx} = (\cos(x))^x \left(-x \tan(x) + \ln(\cos(x))\right) dxdy=(cos(x))x(−xtan(x)+ln(cos(x)))View Full Solution - FreeWas this helpful?