Find dy/dx by implicit differentiation and simplify the result

Question

1. x=sec(1/y)

2. 4xy+ln(x^2y)=7

Hello I sort of know how to do these problems but not really. :( Can someone show all the work to solving them it would help me greatly. Thank you a lot.

Answer 1

x=sec(1/y)
1 = sec(1/y)tan(1/y)(-1/y^2) dy/dx
so,
dy/dx = -y^2 cos(1/y) cot(1/y)

or,

x=sec(1/y)
1/y = arcsec(x)
y = 1/arcsec(x)
y' = -1/[x√(x^2-1) arcsec(x)^2]

you can see that they are the same, since
cos(arcsec(u)) = 1/u
tan(arcsec(u)) = √(u^2-1)

Answer 2

4xy+ln(x^2y)=7
4y + 4xy' + 1/(x^2y)(2xy+x^2y') = 0
4xy' + 1/y y' = -4y - 2/x
y' = -2(2y + 1/x)/(4x + 1/y)
=
-2y(2xy+1)
---------------
x(4xy+1)

or, note that

4xy+ln(x^2y)=7
4xy + 2lnx + lny = 7
4y + 4xy' + 2/x + 1/y y' = 0
(4x+1/y)y' = -4y - 2/x
which works out the same

Answer 3

let z = 1/y

dx = d/dz sec(z)dz
dx = sec (z) tan (z) dz

but dz = (-1/y^2)dy
so
dx = sec(1/y) tan(1/y) (-1/y^2) dy

now
dy/dx = -y^2/[ sec(1/y) tan(1/y]