[SPLIT] implict differentiation for sin(xy) + 2y^3 = 5x^2

[crzymath]

this is another problem: sin(xy)+2y[sup:350tvht1]3[/sup:350tvht1]=5x[sup:350tvht1]2[/sup:350tvht1]

i changed dy/dx to y[sup:350tvht1]1[/sup:350tvht1] to make it easier for me

my work:
sin(xy)+2y[sup:350tvht1]3[/sup:350tvht1]-5x[sup:350tvht1]2[/sup:350tvht1]=0
cos(xy)(xy[sup:350tvht1]1[/sup:350tvht1]+(1)y)+6y[sup:350tvht1]2[/sup:350tvht1]y[sup:350tvht1]1[/sup:350tvht1]-10x=0
xy[sup:350tvht1]1[/sup:350tvht1]cos xy + y cos xy + 6y[sup:350tvht1]2[/sup:350tvht1]y[sup:350tvht1]1[/sup:350tvht1]=10x
y[sup:350tvht1]1[/sup:350tvht1](x cos xy + 6y[sup:350tvht1]2[/sup:350tvht1]) = 10x - y cos xy
y[sup:350tvht1]1[/sup:350tvht1]= 10x - y cos xy/(x cos xy + 6y[sup:350tvht1]2[/sup:350tvht1])
y[sup:350tvht1]1[/sup:350tvht1]= 10x - y/(x+6y[sup:350tvht1]2[/sup:350tvht1]) <---- Reduced answer in simple form; did i do that right?

 

[galactus]

this is another problem: sin(xy)+2y[sup:2ajtl08u]3[/sup:2ajtl08u]=5x[sup:2ajtl08u]2[/sup:2ajtl08u]

i changed dy/dx to y[sup:2ajtl08u]1[/sup:2ajtl08u] to make it easier for me

my work:
sin(xy)+2y[sup:2ajtl08u]3[/sup:2ajtl08u]-5x[sup:2ajtl08u]2[/sup:2ajtl08u]=0
cos(xy)(xy[sup:2ajtl08u]1[/sup:2ajtl08u]+(1)y)+6y[sup:2ajtl08u]2[/sup:2ajtl08u]y[sup:2ajtl08u]1[/sup:2ajtl08u]-10x=0
xy[sup:2ajtl08u]1[/sup:2ajtl08u]cos xy + y cos xy + 6y[sup:2ajtl08u]2[/sup:2ajtl08u]y[sup:2ajtl08u]1[/sup:2ajtl08u]=10x
y[sup:2ajtl08u]1[/sup:2ajtl08u](x cos xy + 6y[sup:2ajtl08u]2[/sup:2ajtl08u]) = 10x - y cos xy
y[sup:2ajtl08u]1[/sup:2ajtl08u]= 10x - y cos xy/(x cos xy + 6y[sup:2ajtl08u]2[/sup:2ajtl08u])
y[sup:2ajtl08u]1[/sup:2ajtl08u]= 10x - y/(x+6y[sup:2ajtl08u]2[/sup:2ajtl08u]) <---- Reduced answer in simple form; did i do that right?

You should get:

\(\displaystyle y'=\frac{10x-ycos(xy)}{xcos(xy)+6y^{2}}\)

Your next to last line is it. I think you reduced a little too much