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

HW

Related Maths A Level answers

All answers ▸

What is the value of sin(theta), cos(theta), tan(theta) where theta = 0, 30, 45, 60, 90


Show that 2(1-cos(x)) = 3sin^2(x) can be written as 3cos^2(x)-2cos(x)-1=0.


Find the derivative for y=5x^3-2x^2+7x-15


Solve the quadratic inequality: x^2 - 5x + 4 < 0