Question

implicit differentiation

Answer 1

\[e^{xy}=\sin (y^2)\]

find y'

Answer 2

ive done it a few different ways and gotten different answers

Answer 3

and the calculators online are also giving me different answers so, would like to see how you guys do it.

Answer 4

\[d(e^{xy}) = d(\sin(y^2)\]Use the chain rule. \[e^{xy} \cdot d(xy) = d(\sin(y^2) \cdot d(y^2)\]

Answer 5

\[
e^{xy} = \sin(y^2) \\
\text{Differentiate with respect to x:} \\
e^{xy} * (xy' + y) = \cos(y^2) * 2y * y' \\
\]Gather up y' terms and solve for y'.

Answer 6

i see

Answer 7

\[\large
y' = \frac{ye^{xy}}{2y\cos(y^2)-xe^{xy}}

\]

Answer 8

\[e^{xy} \cdot( 1\cdot y +xy') = \cos(y^2) \cdot 2yy'\]\[ye^{xy} +xy'e^{xy} =2yy'\cos(y^2)\]\[-xy'e^{xy} +2yy'\cos(y^2)=ye^{xy}\]\[y'(-xe^{xy} +2y\cos(y^2)=ye^{xy}\]\[y'=\frac{ye^{xy}}{-xe^{xy} +2y\cos(y^2)}\]

Answer 9

Eugh, the alignment is off and it looks ugly, but you get the point.

Answer 10

ya i think i could just leave the negative sign out but

Answer 11

looks about right

Answer 12

Why would you leave out the negative sign?

Answer 13

my teacher doesnt care

Answer 14

well one sec

Answer 15

im getting y'=-(e^xy*y)/(e^xy-2ycos(y^2)

Answer 16

or a better way to put it

y'=-(ye^xy)/(e^xy-2ycos(y^2)

Answer 17

same thing thanks

Answer 18

No problem :D

Answer 19

thanks for the medal!