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

16716 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What is the derivative with respect to x of the function f(x)=1+x^3+ln(x), x>0 ?


For the curve, y = e^(3x) - 6e^(2x) + 32 ,find the exact x-coordinate of the minimum point and verify that the y-coordinate of the minimum point is 0.


Differentiate the equation y = (1+x^2)^3 with respect to (w.r.t.) x using the chain rule. (Find dy/dx)


Please explain Pythgoras Theorem


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences