Find the gradient of the curve y = x^2(ln(x)) at x = e

We'll need to use the product rule.

Let's take u = x^2 -> du/dx = 2x, and v = ln(x) -> dv/dx = 1/x

Then dy/dx = x^2(1/x) + 2xln(x) = x + 2xln(x)

Substituting our x value gives (dy/dx)|(x = e) = e + 2eln(e) = 3e

CR
Answered by Charlie R. Maths tutor

5851 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Rewrite ... logF=logG+logH−log(1/M)−2*logR ... in the form F=... using laws of logarithms


Given y = 2x(x^2 – 1)^5, show that dy/dx = g(x)(x^2 – 1)^4 where g(x) is a function to be determined.


How to differentiate y=x^3+4x+1 when x=3


The point A lies on the curve with equation y = x^(1/2). The tangent to this curve at A is parallel to the line 3y-2x=1. Find an equation of this tangent at A. (PP JUNE 2015 AQA)  


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