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.

Answered by Bruno S. Maths tutor

14952 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A curve has equation y = 20x −x2 −2x3 . (A) Find the x-coordinates of the stationary points of the curve.


A ball is fired from a cannon at 20m/s at an angle of 56degrees to the horizontal. Calculate the horizontal distance the ball travels as well as its maximum height reached.


The curve C has the equation: y=3x^2*(x+2)^6 Find dy/dx


What is the integral of sin(3x) cos(5x)?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy