Implicit differentiation in Calculus?

Question

I am really confused and any help would be greatly appreciated thanks.

Find dy/dx by implicit differentiation:
1) 1 + x = sin (xy^2)
dy/dx=?
2) 7x^3 + x^2 y - xy^3 = 4
dy/dx=?

Also if you can help me on this
3) use implicit differentiation to find an equation of the tangent line to the hyperbola @ pt (1,4)

x^2 + 2xy - y^2 + x = -6
y(x)=?

Thank you so much for any help you can give. I have attempted all of them but am unable to get help from my teacher. Please don't just give me the answers, I need an explanation of how to do it.

Answer 1

1) 1 + x = sin (xy^2)
1 = cos(xy^2)(y^2 + 2xyy')
y' = [1-y^2cos(xy^2)]/(2xycos(xy^2))

2) 7x^3 + x^2 y - xy^3 = 4
21x^2 + 2xy + x^2y' - y^3 -3xy^2y' = 0
y' = (21x^2 + 2xy - y^3)/(3xy - x^2)

3) use implicit differentiation to find an equation of the tangent line to the hyperbola @ pt (1,4)
x^2 + 2xy - y^2 + x = -6
2x + 2y + 2xy' - 2yy' + 1 = 0
2 + 8 + 2y' - 8y' + 1 = 0
y' = 11/6
y = (11/6)(x-1) + 4

Answer 2

1)
1 + x = sin(xy^2)

differentiating implicitly

0 + 1 = cos(xy^2)*[x(2yy') + y^2]

1 = 2xy y' cos(xy^2) + y^2 cos(xy^2)

2xy y' cos(xy^2) = 1 - y^2 cos(xy^2)

y' = [1 - y^2cos(xy^2)]/2xy cos(xy^2)

2)

7x^3 + x^2y - xy^3 = 4

differentiating

21 x^2 + x^2 (y') + y(2x) - x(3y^2 y') - y^3 = 0

21x^2 + x^2 y' + 2xy - 3x y^2 y' - y^3 = 0

y'(x^2 - 3x y^2) + 21x^2 + 2xy - y^3 = 0

y'(x^2 - 3x y^2) = - 21x^2 - 2xy + y^3

y' = [y^3 - 21x^2 - 2xy ]/(x^2 - 3xy^2)

3)

To find an equation of tangent to the curve, you should find out the slope first

The slope of tangent is derivative of the curve at the given point

The equation of curve is

x^2 + 2xy - y^2 + x = -6

differentiating

2x + 2x y' + 2y - 2y y' + 1 = 0

=> y'(2x - 2y) + 2x+ 2y + 1 = 0

=>y'(2x - 2y) = - [ 2x+ 2y + 1]

y' = - [ 2x+ 2y + 1] /(2x - 2y)

y' at (1,4)

= -[2 + 8 + 1]/(2 - 8)

=>-(11)/(-6) = 11/6

so the slope of tangent = 11/6

The equation of tangent with slope 11/ 6 and going through

the point (1,4) is

y - 4 = 11/6(x - 1)

6y - 24 = 11x - 11

6y = 11x + 13

Answer 3

1) 1 + x = sin (xy^2)
dy/dx=?

Answer:
0+1=(y^2+2xyy')cos(xy^2)



2) 7x^3 + x^2 y - xy^3 = 4
dy/dx=?
Answer:

21x^2+2xy+x^2y'-(y^3+3xy^2y')=0

arrange to find y'=?





3) use implicit differentiation to find an equation of the tangent line to the hyperbola @ pt (1,4)

x^2 + 2xy - y^2 + x = -6
y(x)=?

take derivative
2x+2y+2xy'-2yy'+1=0
x=1
y=4
2+8+2y'-8y'+1=0
11=6y'

y'=11/6

tangent equation
y-4=11/6(x-1)