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. 

Answered by Kalid U. Maths tutor

6595 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

Let f(x)=x^2-ax+a-1 and g(x)=x-5. The graphs of f and g intersect at one distinct point. Find the possible values of a.


Solve: 1/3 x = 1/2 x + (− 4)


How do I derive the indefinite integral of sine?


f(x)=(2x+1)^0.5 for x >-0.5. Find f(12) and f'(12)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences