Urgent calculus and trig help?

Question

Can you explain and give the answer to these questions

Use implicit differentiation to solve

(a) xy + x + y = x^2y^2

(b) x^3 + 3x2y + y^3 = 8

Is sec inverse of - square root of 2
a) pi/4
b) 3pi/4
c) 5pi/4

THANKS

the question (b) of implicit differentiation is
Use implicit differentiation to find dy/dx.

x^3 + 3x^2y + y^3 = 8

Answer 1

a) xy + x + y = (x^2)(y^2)

Differentiate both sides, keeping in mind that whenever you differentiate y with the chain rule, you express it as dy/dx.

y + x(dy/dx) + 1 + (dy/dx) = (2x)(y^2) + (x^2)(2y)(dy/dx)

Move everything with a dy/dx to the left hand side, and everything else goes to the right hand side.

x(dy/dx) + (dy/dx) - (x^2)(2y)(dy/dx) = (2x)(y^2) - y - 1

Factor dy/dx,

(dy/dx) [ x + 1 - (x^2)(2y) ] = (2x)(y^2) - y - 1

(dy/dx) [ x + 1 - 2(x^2)y ] = (2xy^2 - y - 1)

Divide both sides to isolate dy/dx.

dy/dx = [2xy^2 - y - 1] / [x + 1 - 2x^2 y]

2) x^3 + 3(x^2)y + y^3 = 8

Differentiate implicitly.

3x^2 + (6x)(y) + (3x^2)(dy/dx) + 3y^2 (dy/dx) = 0

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

(dy/dx) [ 3x^2 + 3y^2 ] = [-3x^2 - 6xy ]

dy/dx = [ -3x^2 - 6xy ] / [3x^2 + 3y^2]

Factor top and bottom.

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

Cancel the 3's,

dy/dx = [ (-x)(x + 2y) ] / [x^2 + y^2]

3) sec^(-1)(sqrt(2))

First, remember what interval secant inverse is defined on.
If y = sec^(-1)(sqrt(2)), then

sec(y) = sqrt(2) on the interval 0 ≤ y < π/2 or π/2 < y ≤ π.

If we take the reciprocal of both sides,
cos(y) = 1/sqrt(2), which means
y = pi/4 (since it falls in the above interval).

Answer 2

For implicit differentiation, you just have to remember that you are taking the derivative with respect to x, not y. y is a function of x, and so its derivative is dy/dx, which is important to know when using chain rule.

(a) From the product rule, the derivative of xy is x*(dy/dx) + y.
The derivative of x is 1 and the derivative of y is dy/dx,
The derivative of the left hand side is then:
x*(dy/dx) + y + 1 + dy/dx

The derivative of the right hand side is from the product rule:
2xy^2 + 2(x^2)y*(dy/dx). Do not forget to apply chain rule here.

Setting the two sides equal to eachother again,

x*(dy/dx) + y + 1 + dy/dx = 2xy^2 + 2(x^2)y*(dy/dx)

Now you rearrange your terms so that every term with dy/dx is on one side of the equation and everything else is on the other:
x*(dy/dx) + dy/dx - 2(x^2)y*(dy/dx) = 2xy^2 - y - 1

Now we factor out the dy/dx, obtaining only 1 copy of it, and divide both sides by what is left, to get dy/dx = f(x,y)
dy/dx *(x + 1 - 2(x^2)y) = 2xy^2 - y -1
dy/dx = (2xy^2 - y - 1) / (x + 1 - 2(x^2)y).

(b) Is very similar, but typed a little ambiguously, so I'll leave that one.

Last one:
sec(x) = 1/cos(x), so we want an angle x such that:
1/cos(x) = sqrt(2).
=> cos(x) = 1/sqrt(2).
arcsec(x) has a range from [0, pi], so we need an angle in that range. The only angle that satisfies both of these conditions (not even only among the ones listed, but out of any angle) is pi/4.

Answer 3

Question a)
xy + x + y = x².y²
y + x.dy/dx + 1 + dy.dx = 2x.y² + 2y.dy/dx.x²
(x + 1 - 2.y.x²).(dy/dx) = 2.x.y² - 1 - y
dy/dx = (2.x.y² - 1 - y) / (x + 1 - 2.y.x²)

Question b)
3x² + 6y.dy/dx + 3y².dy/dx = 0
(3y² + 6y).dy/dx = - 3.x²
dy/dx = - 3.x² / [ (3y).(y+2) ]
dy/dx = - x² / [ y (y + 2) ]

Are we saying here that sec x = - √2 ?
if so:-
1/cos x = - √2
cos x = - 1/√2
This means that angle is in 2nd and third quadrants
ie x = 135° and 225°
ie x = 3π/4 and 5π/4

If secx = √2, angle will be in 1st and 4th quadrants
ie x = π/4 and 7π/4