The curve C has the equation y=(15−x)lnxy = (15 - x) \ln xy=(15−x)lnx, x>0x > 0x>0. C has a stationary point at P and dydx=15x−1−lnx\displaystyle \frac{dy}{dx} = \frac{15}{x} - 1 - \ln xdxdy=x15−1−lnx
Show that the x x\,x coordinate of P is a solution of x=151+lnx\displaystyle x = \frac{15}{1 + \ln x}x=1+lnx15
329 exam-style questions on Edexcel A Level Maths Numerical Methods, covering 10.1 Locating Roots, 10.2 Iteration, 10.3 The Newton-Raphson Method, and 10.4 Applications to Modelling. Each one has a worked solution and a mark scheme showing where the marks go.