Calculus Question regarding implicity dif + tangent?

Question

The slope of the tangent of a curve y^3*x+y^2*x^2=6 at (2,1) is?

Answer 1

y^3dx/dx+3y^3xdy/dx+2y^2xdx/dx+2yx^2dy/dx=0
so dy/dx=-(y^3-2y^2x)/(3y^3x+2yx^2)
at (2,1) dy/dx=3/14

Answer 2

First calculate f(2) f(2) = (4-3)^(5/2) = a million . . . . . . tangent and well-known lines bypass by element (2, a million) f'(x) = 5/2 (2x-3)^(3/2) * 2 f'(x) = 5 (2x-3)^(3/2) Now calculate f'(2) f'(2) = 5 (4-3)^(3/2) = 5 Slope of tangent line: 5 Slope of standard line -a million/5 Equation of tangent line, slope = 5, element (2,a million) y - a million = 5 (x - 2) y = 5x - 10 + a million y = 5x - 9 Equation of standard line, slope = -a million/5, element (2,a million) y - a million = -a million/5 (x - 2) y = -a million/5 x + 2/5 + a million y = -a million/5 x + 7/5

Answer 3

To differentiate implicitly, differentiate just as you normally would both x's and y's. The big difference is put dy/dx next to each y you differentiate and treat multiplications as an application of the product rule. Here we go:

3y^2 * dy/dx *x+y^3+ 2y*x^2*dy/dx+2x*y^2=0

Now factor out dy/dx and make it the subject of the formula to get

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

Now just substitute the values of x and y to get dy/dx.