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

4437 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Determine the integral: ∫x^(3/4)dx


For rectangles of area 100 m^2 what is the perimeter of the rectangle with the smallest perimeter?


differentiate with respect to x: (x^3)(e^x)


Given ∫4x^3+4e^2x+k intergrated between the bounds of 3 and 0 equals 2(46+e^6). Find k.


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