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

3071 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given that x = 4sin(2y + 6), Find dy/dx in terms of x


The polynomial f(x) is define by f(x) = 3x^3 + 2x^2 - 8x + 4. Evaluate f(2).


Show that the curve with equation y=x^2-6x+9 and the line with equation y=-x do not intersect.


Differentiate 4(x^3) + 3x + 2 with respect to x


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences