Question

Implicit differentiation help

Answer 1

\[e^ycosx = 1+\sin(xy)\]

Answer 2

have you tried it ?
you'll need product rule and chain rule.
know them ?

Answer 3

yeah i have finished it, just need to check my answer.

I used the product rule initially:

\[\frac{ d }{ dx }e^y.cosx + e^y.\frac{ d }{ dx }cosx\]

Answer 4

resulting in:

\[e^y.\frac{ dy }{ dx }.cosx + e^y.(-sinx) = 0 + \cos(xy).(y+x.\frac{ dy }{ dx })\]

Answer 5

going good till now, correct.

Answer 6

continuing:

\[e^y.cosx.\frac{ dy }{ dx } - e^ysinx = y.\cos(xy) +x.\cos(xy).\frac{ dy }{ dx }\]

--> distributed right hand side by cos(xy)

Answer 7

would that be correct to distribute like i did?

Answer 8

yes, go on

Answer 9

collect the terms with dy/dx together

Answer 10

continuing:

\[e^y.cosx.\frac{ dy }{ dx } - x.\cos(xy).\frac{ dy }{ dx }=y.\cos(xy)+e^y.sinx\]

--> moved dy/dx to the left.

Answer 11

correct.

Answer 12

continuing: take out common dy/dx on the left:

\[\frac{ dy }{ dx }(e^y.cosx- x.\cos(xy))=y.\cos(xy)+e^y.sinx\]

divide right side by factor of the left side:

\[\frac{ dy }{ dx }=\frac{y.\cos(xy)+e^y.sinx}{(e^y.cosx- x.\cos(xy))}\]

Answer 13

thats correct, i got the same.
good work! :)

Answer 14

thanks so much!

Answer 15

welcome ^_^