Question

Find dy/dx when ln xy- 3x=5

Answer 1

start with
\[\frac{1}{xy}\times (y+xy')-3=0\] and solve for \(y'\)

Answer 2

or if you prefer, since \(\log(xy)=\log(x)+\log(y)\) you can start with
\[\log(x)+\log(y)-3x=5\] then get
\[\frac{1}{x}+\frac{y'}{y}-3=0\] and solve for \(y'\)
may be quicker

Answer 3

The answer choices all include e^3x+5 which I dont understand.

Answer 4

The choices are
1)dy/dx= ((3x-1)/x)e^(3x+5)

2) dy/dx = ((3x-5)/x)e^(3x+5)

3) dy/dx = ((3x-1)/x^2)e^(3x+5)

4) dy/dx = ((1-3x)/x^2)e^(3x+5)

5)dy/dx = ((5-3x)/x)e^(3x+5)

6) dy/dx = ((3x-5)/x^2)e^(3x+5)

It is not 1

Answer 5

@saw3266
Can you show your working ??

Answer 6

what?

Answer 7

Did you use satellite's method ??

Answer 8

I got y'= -y/x +3y but that doesnt have e^(3x+5) in it

Answer 9

Can you find the value of y from the original quesiton ??

Answer 10

it is
\[\ln(xy)-3x=5\] right? or did i read it wrong?

Answer 11

i guess you could start with
\[\ln(xy)=3x+5\] and so
\[xy=e^{3x+5}\] and go from there

Answer 12

take the derivative of both sides, get
\[xy'+y=3e^{3x+5}\] then solve for \(y'\)

Answer 13

opps. I made a mistake with the answer choices. 2,4,6 are e^3x-5

Answer 14

OHHH! that make sense! thanks!

Answer 15

@satellite73

Answer 16

i didnt realize the whole ln to e thing. Our teacher is ridiculously smart and skips steps like that.

Answer 17

Please help me with this satellite73