A curve CCC has equation
y=xcosxx>0,y>0 y = x^{\cos x} \quad x > 0, \quad y > 0 y=xcosxx>0,y>0Find, by firstly taking natural logarithms, an expression for dydx\frac{dy}{dx}dxdy in terms of xxx and yyy.
Hence show that the xxx-coordinates of the stationary points of CCC are solutions of the equation
sin(x)⋅xlnx=cosx \sin(x) \cdot x \ln x = \cos x sin(x)⋅xlnx=cosx649 exam-style questions on Edexcel A Level Maths Differentiation, covering 9.1 Differentiating sin x and cos x, 9.2 Differentiating exponentials and logarithms, 9.3 The Chain Rule, 9.4 The Product Rule, 9.5 The Quotient Rule, 9.6 Differentiating Trigonometric Functions, 9.7 Parametric Differentiation, 9.8 Implicit Differentiation, 9.9 Using Second Derivatives, and 9.10 Rates of Change. Each one has a worked solution and a mark scheme showing where the marks go.