Question

implicit differentiation y^2 +4xy +x^2 = 13... i need to know how the 4xy gets differentiated with product rule

Answer 1

4y+4x(dy/dx)

Answer 2

not correct.. i am pretty sure you are wrong

Answer 3

think of \(y=f(x)\) so every place you see a \(y\), mentally replace it by \(f(x)\)

Answer 4

i have (4) (xy) so.... 4' times xy + 4 times xy'

Answer 5

\[y^2 +4xy +x^2 = 13..\]
\[f^2(x)+4xf(x)+x^2=13\]
\[2f(x)f'(x)+4(f(x)+xf'(x))+2x=0\] more easiliy written as
\[2yy'=4(y+xy')+2x=0\]

Answer 6

lol, if anything

[4xy]' = 4'xy + 4x'y + 4xy'

Answer 7

the 4 is a constant, so treat it like one
you don't use the product rule when you take the derivative of \(4\sin(x)\) you just write \(4\cos(x)\) right?

Answer 8

you can do the product rule on it; the 4' part goes 0 tho

Answer 9

the product rule is needed because you have \(x\times y\) not because of the 4

Answer 10

i dont know .. im still confused.. im having a moment.. ive done these before.. but i dont understand right now

Answer 11

ok suppose you had this:
\[x\sin(x)\] what would you do to find the derivative?

Answer 12

you have to use the product rule, because it is a product. you have to write
\[\sin(x)+x\cos(x)\]

Answer 13

now suppose instead you had
\[4x\sin(x)\]
the 4 doesn't bother you at all, it is still a constant, so the derivative would be
\[4(\sin(x)+x\cos(x))\] the product rule is still necessary, but not because of the 4, but because you have x times sine

Answer 14

now instead of \(x\sin(x)\) you have \(xy\) where \(y\) is some unknown function of \(x\) a

Answer 15

product rule still applies

Answer 16

but the 4 is just sitting there, so the derivative of \(4xy\) with respect to \(x\) is \[4(y+xy')\]

Answer 17

so 4xy' is 4y +4xy'

Answer 18

yes that is the deriviative, right

Answer 19

hope that is more or less clear

Answer 20

its better now.. thanks

Answer 21

yw