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

3071 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The finite region S is bounded by the y-axis, the x-axis, the line with equation x = ln4 and the curve with equation y = ex + 2e–x , (x is greater than/equal to 0). The region S is rotated through 2pi radians about the x-axis. Use integration to find the


A curve has the equation, 6x^2 +3xy−y^2 +6=0 and passes through the point A (-5, 10). Find the equation of the normal to the curve at A.


How do you find the possible values of cos(x) from 5cos^2(x) - cos(x) = sin^2(x)?


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


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy