How do you find the slope of a tangent line to the graph of the function #x^3 + y^3 – 6xy = 0#, at (4/3, 8/3)?

Question

Emilia Aguilar · Accepted Answer

#4/5#

use implicit differentiation

#d/dx(x^3+y^3-6xy)=d/dx(0)#

#3x^2+3y^2*(dy)/dx-6(1)y-6x*(dy)/dx=0#

solve for #(dy)/dx#, which gives the tangent slope at point #(x,y)#

#3y^2*(dy)/dx-6x*(dy)/dx=-3x^2+6y#

#(dy)/dx(3y^2-6x)=-3x^2+6y#

#(dy)/dx=(-3x^2+6y)/(3y^2-6x)#

plug in #(4/3,8/3)#:

#(dy)/dx# at #(4/3,8/3) =(-3(4/3)^2+6(8/3))/(3(8/3)^2-6(4/3))#

#(dy)/dx# at #(4/3,8/3) =(-16/3+16)/(64/3-8)#

#(dy)/dx# at #(4/3,8/3) =(32/3)/(40/3)#

#(dy)/dx# at #(4/3,8/3) =4/5#

Emily Ammerman · Answer

undefined To find the slope of a tangent line to the graph of a function at a specific point, we can use the derivative of the function.

First, we need to find the derivative of the function. Taking the derivative of x^3 + y^3 – 6xy = 0 with respect to x, we get 3x^2 + 3y^2(dy/dx) - 6y - 6x(dy/dx) = 0.

Next, we can substitute the given point (4/3, 8/3) into the equation and solve for dy/dx. Plugging in the values, we have 3(4/3)^2 + 3(8/3)^2(dy/dx) - 6(8/3) - 6(4/3)(dy/dx) = 0.

Simplifying the equation, we get 4 + 64(dy/dx) - 16 - 32(dy/dx) = 0.

Combining like terms, we have 32(dy/dx) + 48 = 0.

Solving for dy/dx, we get dy/dx = -48/32 = -3/2.

Therefore, the slope of the tangent line to the graph of the function x^3 + y^3 – 6xy = 0 at the point (4/3, 8/3) is -3/2.