Find the exact value of the gradient of the curve y = e^(2- x)ln(3x- 2). at the point on the curve where x = 2.

To solve this problem, we must first differentiate:

Identify that we are able to use the product rule as our expression is of the form y = f(x)g(x) where f(x) = e^(2- x) and g(x) = ln(3x- 2). 

Hence f'(x) = -e^(2- x) and g'(x) = 3/(3x- 2)

By the product rule, dy/dx = f(x)g'(x) + f'(x)g(x) = 3e^(2- x)/(3x- 2) - e^(2- x)ln(3x- 2).

When we substitute x = 2 into this equation, we get that dy/dx = 3/4 - ln(4), which is our final answer.

JC
Answered by Joe C. Maths tutor

9685 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

integrate function (x^4+3x)/(x^2) with respect to x


Given that x = ln(sec(2y)) find dy/dx


Tom drink drives two days a week, the chance of him being caught per day is 1 in 100. What is the chance he will not be driving after a) one week? b) one year?


Find the gradient of the tangent to the curve y=4x^2 - 7x at x = 2


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning