Help with a couple of Calculus Problems?

I'll give #2 a try here.

I take the derivative as

((x + 9y)y' - y(1 + 9y'))/(x + 9y)^2 = 7x^6.

(x + 9y)y' - y(1 + 9y') = 7x^6(x + 9y)^2.

y' = (y + 7x^6(x + 9y)^2)/x.

Plugging in the values, I get...

((-8/71) + 7(1 - 72/71)^2) = -561/5041.

I get the answer as -561/5041.

from 12 to 7, is a 7hour difference

ship A, 18x7= 126 nautical miles

ship B, 21x7= 147 nautical miles

draw a triangle, and since ship A started out 50 nautical miles due west of ship B, the length of that distance of 7pm is 176

your triangle, base=x=173, height=y=147, and the distance connecting ship A and ship B =

c^2= 173^2 + 147^2

c^2= 29929 + 21609

c^2= 51538

c=sqrt 51538

x^2 + y^2 = c^2

2x dx/dt + 2y dy/dt = 2c dc/dt

2(173)(18) + 2(147)(21) = 2(sqrt51538)dc/dt

6228 + 6174 = 2(sqrt51538) dc/dt

dc/dt = 6201 / (sqrt51538)

2. first multiple by x+9y

y=(x^7 + 7)(x+9y)

y=x^8 +9yx^7 + 7x + 63y

dy/dx= 8x^7 + 63yx^6 + 9x^7 dy/dx + 7 + 63 dy/dx

-8x^7 - 63yx^6 - 7= 9x^7 dy/dx + 63 dy/dx - dy/dx

-8x^7 - 63yx^6 - 7= 9x^7 dy/dx + 62 dy/dx

-8x^7 - 63yx^6 - 7= dy/dx [9x^7 + 62]

dy/dx = [-8x^7 - 63yx^6 - 7]/ [9x^7 + 62]

plug in x=1, y= 8/-71 and that will be your slope

[-8 + (504/71) - 7] / [9+62]

[-15 + (504/71)] / 71

[ (-1065/71) + (504/71)] / 71

[-561 / 71] / 71

m= -561/ 5041