A curve has equation (x+y)^2=x*y^2, find the gradient of the curve at a point where x=1

  1. Differentiating left hand side: 2(x+y)(1+dy/dx) from the chain rule 2. Differentiating right hand side: y2+2xy(dy/dx) from the product rule 3. Equating sides and taking out factors of dy/dx to rearrange for dy/dx: dy/dx=[y2-2(x+y)]/[2(x+y)-2xy] 4. Substitute x=1 into original expression and solving for y (i.e. solving (1+y)2=y2) gives y=-1/2 5. Substituting x=1 and y=-1/2 into the expression for dy/dx gives dy/dx=-3/8
PK
Answered by Peter K. Maths tutor

4427 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

differentiate 4x^3 + 3x


A curve has parametric equations -> x = 2cos(2t), y = 6sin(t). Find the gradient of the curve at t = π/3.


The polynomial p(x) is, p(x)= x3-5x2-8x+48.Use the Factor Theorem to show that (x + 3)is a factor of p(X)


Find the stationary points on y = x^3 + 3x^2 + 4 and identify whether these are maximum or minimum points.


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:

© 2026 by IXL Learning