The equation of a curve is x(y^2)=x^2 +1 . Using the differential, find the coordinates of the stationary point of the curve.

Firstly we need to use product rule to find the dy/dx of the left hand side (LHS). Using implicit differentiation, we know the differential of y^2 is 2y(dy/dx). Then use to product rule to obtain the dy/dy of LHS to be 2xy(dy/dx). The right hand side, we can treat as a normal differential therefore it is 2x. We can then rearrange the equation so that (dy/dx) is the subject. Now, we need to find the stationary point and to do that, we must set the differential equal to zero and rearrange to get either x or y on its own. I suggest trying to isolate y since it makes the next part a little easier. After rearranging, you should get y=root2x so then we can substitute root2x into the original equation to get the x coordinate. This is 1. To obtain the y coordinate, simply sub 1 into our equation for y and we get +/- root2.

GC
Answered by Grace C. Maths tutor

6055 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A curve has the equation y = x^4 - 8x^2 + 60x + 7. What is the gradient of the curve when x = 6?


Show that the derivative of tan(x) is sec^2(x), where sec(x) is defined as 1/cos(x). [Hint: think of tan(x) as a quotient of two related functions and apply the appropriate identity]


How do I find the equation of the tangent of a curve at a specific point.


Find the derivative (dy/dx) of the curve equation x^2 -y^2 +y = 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:

© 2025 by IXL Learning