Answers>Maths>IB>Article

Given 2x^2-3y^2=2, find the two values of dy/dx when x=5.

First solve for the exact point on the line by substituting 5 into the original equation. You should get y=+-4. 
Now implicitly differentiate the equation: 4x-6y(dy/dx)=0. Rearranging this will yield the following: dy/dx=(2x)/(3y). Because we only have one value of x, let's substitute this into the derivative first: dy/dx=10/3y. Now we can individually substitute the two y values to get the two values of dy/dx.  dy/dx = 10/12 = 5/6, dy/dx = -10/12 = -5/6 These are the two values of dy/dx when x=5. 

KU
Answered by Kalid U. Maths tutor

6865 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

In Topic 5 (Statistics and Probability) what is the difference between mutually exclusive events and independent events?


In an arithmetic sequence, the first term is 2, and the fourth term is 14. a) Find the common difference, d. b) Calculate the sum of the first 14 terms, S14.


What is the equation of the tangent drawn to the curve y = x^3 - 2x + 1 at x = 2?


How do I derive the indefinite integral of sine?


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