Find the tangent of the following curve, y=xe^x, at x=1 expressing it in the form y=mx+c?

Firstly, we calculate the y-value when x=1, namely y=e. Then we need to find the gradient of this curve at x=1, which can be determined by taking the derivative of y and then valuate it at x=1. So dy/dx=xe^x+e^x=(x+1)e^x, at x=1 dy/dx=2e. Using the equation of a line given by y-y_0=m(x-x_0), where m is the gradient of the line (namely m=2e) and (x_0,y_0) is the coordinate that is given to us (namely x_0=1 and y_0=e), we obtain that y-e=2e(x-1), hence y=2ex-e is the tangent of this curve at x=1.