The curve C has equation (4x^2-y^3+3^2x)=0. The point P (0,1) lies on C: what is the value of dy/dx at P?

Use the chain rule to differentiate the original equation: this results in 8x-3y^2*(dy/dx) + 2ln(3)3^2x=0. This can be rearranged to find dy/dx as a function of y and x: 3y^2(dy/dx)=8x+2ln(3)*3^2x -> dy/dx=(8x+2ln(3)3^2x)/3y^2. At this point, dy/dx at point P can be computed: dy/dx=(80+2ln(3)3^0)/31^2=2ln(3)/3

TD
Answered by Tutor65063 D. Maths tutor

3431 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I differentiate and integrate powers of x?


Find and classify all the stationary points of the function f(x) = x^3 - 3x^2 + 8


Factorise the following: 5a^3b^5-4ab^2


integrate x^2(2x - 1)


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:

© 2026 by IXL Learning