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

2886 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the equation of the straight line perpendicular to 3x+5y+6=0 that passes through (3,4)


Integrate dy/dx = 2x/(x^2-4)


Curve C has equation x^2 - 3xy - 4y^2 + 64 = 0. a) find dy/dx in terms of x and y. b) find coordinates where dy/dx=0.


How would you sketch the curve of a graph?


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
Cookie Preferences