Given the equation 0=5x^2+3xy-y^3 find the value of dy/dx at the point (-2,2)

To answer this we will use implicit differentiation with respect to x. So start by differentiating each term. On the left hand side 0 differentiates to 0. On the right hand side 5x2 differentiates to 10x. By using the product rule and implicit differentiation 3xy differentiates to 3x dy/dx +3y. -y3 differentiates to -3y2 dy/dx by implicit differentiation. So the whole differentiated equation is 0=10x+3x dy/dx +3y - 3y2 dy/dx. Then rearrange the equation so all terms containing dy/dx are on one-side of the equals sign and the other terms are on the other-side so 3y2 dy/dx -3x dy/dx = 10x+3y. Then take out a factor of dy/dx from the left hand side giving dy/dx(3y2-3x)=10x+3y. Finally, divide each side by 3y2-3x to get an equation in terms of dy/dx, dy/dx=(10x+3y)/(3y2-3x). Then plug in the co-ordinates given above to obtain dy/dx=-7/9

Answered by Holly W. Maths tutor

3243 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A curve has equation y = x^3 - 3x^2 -24x + 5, find the x co-ordinates of the two stationary points of the curve and hence determine whether they are maximum or minimum points.


Express 4x/(x^2-9) - 2/(x+3) as a single fraction in its simplest form.


Show that the derivative of tan(x) is sec^2(x), where sec(x) is defined as 1/cos(x). [Hint: think of tan(x) as a quotient of two related functions and apply the appropriate identity]


A car is accelerating at 2 ms^-2 along a horizontal road. It passes a point A with a velocity of 10 ms^-1 and later a point B, where AB = 50m. FInd the velocity of the car as it passes through B.


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences