Find the an expression for dy/dx of the function y=(4x+1)ln(3x+1) and the gradient at the point x=1.

This question is example of product rule. To simplify need to define two terms u=4x+1 and v=ln(3x+1). Each term is then differentiated. du/dx=4. To differentiate v, need to define another term a=3x+1, therefore v=ln(a). Differentiate using chain rule, dv/dx=da/dxdv/da, therefore dv/dx=3/(3x+1). Product rule then states dy/dx=udv/dx+vdu/dx, therefore, dy/dx=(4x+1)(3/(3x+1))+4(ln(3x+1))dy/dx=(12x+3)/(3x+1)+4ln(3x+1)To find the gradient at the point x=1, need to substitute into dy/dx.dy/dx(x=1)=(12+3)/(3+1)+4ln(3+1)=15/4+4ln4=9.295