A curve has equation y^3+2xy+x^2-5=0. Find dy/dx.

To find the derivative of this, we must differentiate each term with respect to x. This implies d/dx(y^3+2xy+x^2-5=0). We can differentiate each term seperately so d/dx(y^3+2xy+x^2-5=0) is equal to d/dx(y^3) + d/dx(2xy) +d/dx(x^2) - d/dx(5) = 0. Taking each term seperately, d/dx(y^3) = (dy/dx)(d/dy(y^3) = dy/dx(3y^2), d/dx(2xy) = 2y+dy/dx(d/dy(2xy)) = 2y+2x(dy/dx), d/dx(x^2) = 2x, d/dx(5) = 0. Recombining we get dy/dx(3y^2)+2y+dy/dx(2x)+2x=0. Rearranging and factorising gives us dy/dx(3y^2+2x)=-(2x+2y). Dividing by 3y^2+2x then gives us dy/dx = -(2x+2y)/3y^2+2x. ARQ. 

MC
Answered by Matthew C. Maths tutor

4738 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The curve C has equation 4x^2 – y^3 – 4xy + 2^y = 0 The point P with coordinates (–2, 4) lies on C . Find the exact value of dy/dx at the point P .


What is the gradient of the function f(x) = 2x^2 + 3x - 7 at the point where x = -2?


Use the substitution u = 2^x to find the exact value of ⌠(2^x)/(2^x +1)^2 dx between 1 and 0.


A curve has equation y = x^3 - 6x^2 - 15x. The curve has a stationary point M where x = -1. Find the x-coordinate of the other stationary point on the curve.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences