Question

Help!!! Find dy/dx : a. y= log(base2) (3x) b. x^2y^4 = ln x + 3y

Answer 1

For the part "a" can I use Chain Rule to solve ??

Answer 2

b.
x^2 4y^3 * dy/dx + 2x y^4 = 1/x + 3 * dy/x

solve for dy/dx

Answer 3

for a. i think you need to the log to base e

Answer 4

i'm having a hard time remembering formula for change of base

try googling it

Answer 5

* convert the log to base e

Answer 6

Me either.....I don't remember the formula.

Answer 7

On part b. which term I will divide on both side ???

Answer 8

\[ y' = (\log_2{3x})'\\\frac{dy}{dx}=\frac{3}{xln2}\]

Answer 9

sorry, RHS doesn't have 3 in numerator

Answer 10

okay.....than will be dy/dx= x ln2 ??

Answer 11

1/xln2

Answer 12

ok...thank you Loser 66. I appreciate it !!!!

Answer 13

in part b isolate all the terms containing dy/dx
then factor dy/dx out and then divide

Answer 14

In part b, I think it's implicit differentiate .

Answer 15

like what cwrw238 said.

Answer 16

I am trying........

Answer 17

Hello guys.....I still need help on part b.

Answer 18

\[x^2y^4=lnx+3y\\x^2y^4-3y=lnx\]
derivative both sides
\[x^2*4y^3*y' -3y'=\frac{1}{x}\]
factor y' out
\[y'(4x^2y^3-3) =\frac{1}{x}\\y'=\frac{1}{x(4x^2y^3-3)}\]
or \[\frac{dy}{dx}= \frac{1}{x(4x^2y^3-3)}\]

Answer 19

Thanks again Loser66.......I tried a lot.....unsuccessfully !!!

Answer 20

yw