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=cosx333 exam-style questions on WJEC A Level Maths 3.6 Differentiation (A-level only), covering 3.6.1 Differentiation (A-level only), 3.6.2 Differentiation (A-level only), 3.6.3 Differentiation (A-level only), 3.6.4 Differentiation (A-level only), 3.6.5 Differentiation (A-level only), 3.6.6 Differentiation (A-level only), 3.6.7 Differentiation (A-level only), and 3.6 Differentiation (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.