Question

Find the indicated derivative:

3y + pisin(x^2 y) = pi at (1, pi/6)

Thanks.
Find the indicated derivative:

3y + pisin(x^2 y) = pi at (1, pi/6)

Thanks.
@Mathematics

Answer 1

\[3y +\pi \sin(x ^{2}y) = \pi \]

Answer 2

(3y)'=3(y)'=3y'
sin(g(x))=g'(x)cos(g(x))
if g(x)=x^2y
then
g'(x)=(x^2y)'=(x^2)'y+x^2(y)'
= 2xy+x^2y'
(constant)'=0

Answer 3

\[3 + \cos[x^2dy/dx + 2xy] = 0 \] by using the product rule I came up with the top ^ and then I simplified and solved for dy/dx though I keep getting the wrong answer, would I be on the right track though?

Answer 4

\[3y'+\pi\cos(x^2y)(2xy+x^2y')=0\]and we solve for y', or else plug in the numbers now and solve for y' second

Answer 5

looks like a mistake in your answer above because you have 3 instead of 3y', and also the pi is missing as a factor of the second term

Answer 6

probably easiest now to replace x by 1 and y by
\[\frac{\pi}{6}\] and see what we get.

Answer 7

\[3y'+\pi\cos(\frac{\pi}{6})(\frac{\pi}{3}+y')=0\]

Answer 8

now algebra. hope it works

Answer 9

Ohhh okay, alright I think I get it now, thanks a lot for your help. I appreciate it !

Answer 10

yw. i got an answer but it sure looks ugly!

Answer 11

lol hmm the answer should be \[(\pi ^{2}\sqrt{3}) / (18 + 3 \pi \sqrt{3}) \]

Answer 12

this is a long tedious differentiation ...

Answer 13

Yes, I hate these questions, its going to be all over my midterm :S

Answer 14

i got

dy/dx = -2 pi yx cox(yx^3) / pi x^2 cos(yx^2) + 3

Answer 15

ohh any chance you could show me how you did that?

Answer 16

sorry its yx^2 not yx^3

Answer 17

right - i did it on pencil/paper - i'll take another look and try to explain it - it is long winded

Answer 18

alright thanks a lot.

Answer 19

sorry - i've been called away - be back later

Answer 20

np, thanks for your help anyways

Answer 21

3y + pi sin (x^2y) =pi

3 dy/dx + d(pi sin (x^2y) /dx = 0

find d(pi sin (x^2y) /dx:
= (x^2 dy/dx + 2xy) * pi cos(x^2y) - by product / chain rule

= pi cos(x^2y) * x^2 dy/dx + pi cos(x^2y) * 2xy

so
3 dy/dx + pi cos(x^2y) * x^2 dy/dx + pi cos(x^2y) * 2xy = 0
3 dy/dx + pi cos(x^2y) * x^2 dy/dx = - 2 pi xy cos(x^2y)

dy/dx = [ -2pi xy cos (x^2 y)] / [pi x^2 cos(x^2 y) + 3]

Answer 22

value at (1.pi/6):

= -2 pi (pi/6) cos (pi/6) / pi cos (pi/6) + 3

= -2 (pi)^2/6 * sqrt3/2
----------------
sqrt3 pi/ 3 + 3

= - sqrt3 (pi)^2 2
----------- * ------
6 sqrt3 pi + 6

= -sqrt 3 (pi(^2 ) / ( 3 sqrt3 pi + 18)