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
Answered by Peter K. Maths tutor

3526 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do you find the x co-ordinates of the stationary points of a curve with the equation y = 10x - 2x^2 - 2x^3


The height x metres, of a column of water in a fountain display satisfies the differential equation dx/dt = 8sin(2t)/(3sqrt(x)), where t is the time in seconds after the display begins. (a) Solve the differential equation, given that x(0)=0


Given that y = ((4x+3)^5)(sin2x), find dy/dx


y = 4x / (x^2 + 5). Find dy/dx.


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