Find the equation of the tangent line to the curve 3xy = x3+y3, at the point (2/3,4/3). HELP!?

Find the equation of the tangent line to the curve 3xy = x3+y3, at the point (2/3,4/3). HELP!?

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. y^2(y^2 − 4) = x^2(x^2 − 5) at (0, −2)

dx/dt=2,dx/dt=2, and dy/dt=6,dy/dt=6, find dz/dtdz/dt when x=6x=6 and y=8. z^2=x^2+y^2 z=(x^2+y^2)^(1/2) dz/dt= ..I'm stuck at this point how do I differentiate...

I don't get how to do this and I'm not sure how to use the ln y' =

A runner sprints around a circular track of radius 100m at a constant speed of 7m/s.The runner's friend is standing at a distance 200m from the center of track. How fast is the distance between...

If the point (-5,0) is on the edge of the shadow, how far above the x-axis is the lamp located i got the y' (slope) from differentiating the equation and get y'= -x/4y, but i cannot...

x+2y=sqrt(y)

Hi, this one has me got me confused. find dy/dx at the point (0, -1), if y7=(x2-1)3+4extan-1y+pi Any help would be greatly appreciated. &n...

I need help answering this word problem...I need to find the function P(t) and I just can't remember the process whatsoever "A company has its profit function P(t) defined since...

In the middle of solving this implicit function/using implicit differentation to find its tangent line, this is what I have so far: Original function: y^3+x^3-2xy=8 (find tangent...

find dy/dx

Find dy/dx by implicit differentiation : e^(x/y) = x - y

use implicit differentiation

Need to use implicit differentiation.

I'm having trouble solving this problem.

Suppose y is defind implicitly as a function of x by the following equations, where g is a given differentiable function of one variable. Find an expression for y'. a. xy=g(x)+y3 Ans:...

(x2/a2) + (y2/b2)=1 Point (m,n) Answer should be in terms of x,y,m,n,a,b and answer is _______=1

The original equation is: cos(x^2y^2)/sin(x^2y^2)=2x I simplified it to cot(x^2y^2)=2 I used chain rule to get -csc^2u * du/dx and found du/dx= (2x^2yy'+2y^2x) new...

the position function s of a point is given by two dimensional quantities where t in seconds and the y direction is affected by gravity (10m/s/s). y(t)=4t*sin(θ) - 5t2 ...

At what point is the tangent line to the curve y^3=2x^2 perpendicular to the line x-2y-10=0 ? (x,y) format