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

7130 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

Two functions, y1 & y2, are given by y1=x^2+16x+4; y2=2(3x+2). Find analytically the volume of the solid created by revolving the area between the two curves by 2pi radians around the x-axis. N.B. y2>y1 on the interval between the points of intersection.


Determine the integral: ∫5x^4dx


Write down the expansion of (cosx + isinx)^3. Hence, by using De Moivre's theorem, find cos3x in terms of powers of cosx.


Consider the arithmetic sequence 5,7,9,11, …. Derive a formula for (i) the nth term and (ii) the sum to n terms. (iii) Hence find the sum of the first 20 terms.


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