Find y′ in each of the following equations. If 3x^2+11y^2=14, then y′=? If 19xy+cos(14y)=8x, then y′=?

Answer 1

3x²+11y²=14

This is an equation of ellipse. Therefore we expect to have 2 answers for y' in terms of x.

11y² = 14-3x²

y²=(14-3x²)/11

y = ± ((14-3x²)/11)^1/2

Using the original equation, use implicit differentiation:

6x + 22yy' = 0

22yy' = -6x

y' = -6x/(22y) = -3x/(11y)

Plugin the value of y to get the value of y' in terms of x:

y' = -3x/ (±11((14-3x²)/11)^1/2 )

y' = ± 3x/(11(14-3x²)^1/2)

y' = ± 3x/(154-33x²)^1/2

Where -(14/3)^1/2 < x < (14/3)^1/2

19xy+cos(14y)=8x

The graph of this equation has vertical oscillation and as x approaches 0, we are getting more values for y' and each value on the right hand side is increasing greatly and on the left hand side is decreasing greatly as well. You will see what I'm saying if you graph the equation in any graphing utility. Therefore, y' cannot be expressed in terms of x alone. it should be accompanied by y to make it simple.

Using implicit differentiation:

(19xy' + 19y) - 14y'sin (14y) = 8

19xy' - 14y'sin (14y) = 8 - 19y

y'(19x -14 sin (14y)) = 8 - 9y

y' = (8 - 9y)/(19x-14 sin (4y))

Answer 2

3x^2 + 11y^2 = 14

6x + 22yy' = 0

y' = -6x/22y = -3x/11y

y' =(-3/11)(x/y)

3x^2 + 11y^2 = 14

11y^2 = 14 -3x^2

y^2 = 14/11 - 3x^2/11

y = + or - sqr(14/11 -3x^2/11)

it's an ellipse with center at the origin (0,0)

3x^2/14 + 11y^2/14 = 1

x^2/(14/3) + y^2/(14/11) = 1

x^2/(sqr(14/13))^2 + y^2/(sqr(14/11))^2 = 1

the ellipse has semi-axes sqr(14/13) and sqr(14/11)

11y^2 = 14-3x^2

y^2 = 14/11 -3x^2/11

y = + or - sqr((14-3x^2)/11)

y' = (-3/11)(x/y) = (-3/11)(x/(+or-sqr((14-3x^2)/11)

= - or + (3x/11)(sqr((14-3x^2)/11)

or just y' = -3x/11y

y=0, slope is undefined, at the right and left ends of the ellipse

x= 0, slope = 0, the top and bottom of the ellipse

3x^2 + 11y^2 = 14

for x = sqr(14/3)

3(14/3) + 11y^2 = 14

11y^2 = 0, y= 0. that's the right & left end points of the ellipse

where y'= -3x/11(0) is undefined.

Answer 3

IMPLICIT DIFFERENTIATION MADE EASY USING PARTIAL DERIVATIVES

CASE I

Let us think that the equation 3 x² + 11 y² =14 defines y implicitly as a differentiable function of x., that is

y = f(x) , where F ( x, f(x) ) = 14 for all x in the domain of f.

Then d y / d x = - [ ∂ F / ∂ x ] / [ ∂ F / ∂ y ] = - 6 x / 22y = - 3 x / 11 y.

CASE II

F( x , y ) = 19 x y + cos ( 14 y) -8 x = 0

As before d y / d x = - [ ∂ F / ∂ x ] / [ ∂ F / ∂ y ] = - [ 19 y - 8 ] / [ 19 x - 14 sin ( 14 y ) ]

= [ 8 - 19 y ] / [ 19 x - 14 sin ( 14 y ) ]

Hence d y / d x = [ 8 - 19 y ] / [ 19 x - 14 sin ( 14 y ) ]