using implicit differentiation how do you find the slop of the tangent line to the curve xy^3+xy=8?

Question

at the point (4,1)

Answer 1

xy^3 + xy=8

Step 1: Rewrite so that x and y are on opposite sides.
y^3 + y = 8/x = 8x^-1

Step 2: Differentiate both sides.
3y^2 + 1 dy = -8x^-2 dx = -8/x^2 dx
dy / dx = (-8/x^2) / 3y^2 + 1
dy / dx = -8 / (x^2)(3y^2 + 1)

Step 3: Plug in x and y to find slope (because the derivative of the function is used to find slope):
dy/dx = -8 / (4^2)(3(1^2) + 1)
dy/dx = -8 / (16)(4)
dy/dx = -8 / 64 = -1/8 (solution!)

Answer 2

i will take it one after the different. Differentiating xy^3 implicitly: x (3y^2)(d/dx) + y^3 = 3xy^2(d/dx) + y^3 Differentiating xy implicitly: x (d/dx) +y at the same time as 8 differentiates to 0. the hot equation is: 3xy^2(d/dx) + y^3 + x (d/dx) +y =0 Factorizing: d/dx (3xy^2 + x) = -y -y^3 d/dx= ( -y -y^3) / (3xy^2 + x) At (4, a million), d/dx= (-a million-a million)/(3×4×a million + 4) = -2/sixteen = -a million/8 (m is -a million/8 ) Eqn of tan line: y-a million = -a million/8 (x-4) Y= -x/8 + a million/2 + a million= -x/8 + 3/2 (b is 3/2) Simplifying further: 8y = 4 - x (eqn of tan line) wish that facilitates!