Question

need help and show ur solution so i can understand it
use implicit differentiation

Answer 1

\[x(x+y)=x ^{4}y ^{2}-y\]

Answer 2

ok, try to do what comes naturally .. what do you come up with?

Answer 3

you know product rule and power rule .... those dont change

Answer 4

plss show how u solve it . im confused in this problem

Answer 5

i need to see your confusion to know where it resides. show me what you think should happen ....


lets start with one piece of it, you see its a product .... what should we apply?

x(x+y)

Answer 6

\[x ^{2}+xy\]???

Answer 7

i really dont know

Answer 8

wasnt what i was going for, but that is a correct move nonetheless

so, what is the d/dx of x^2?

Answer 9

what do you normally do when you see x^2 ?

Answer 10

2x

Answer 11

good

now, is that other term, xy, a product?

Answer 12

no????

Answer 13

xy means x times y and a product is math vocabulary for: "we multiply things together"

Answer 14

xy is a product of x times y, therefore it IS a product

what is the product rule for derivatives?

Answer 15

xy

Answer 16

spose we have to functions f(x) and g(x); or simply f and g

the derivative of fg would simply be f'g + fg'

does this ring a bell?

Answer 17

xy + xy' ???????

Answer 18

very close :) you prolly missed the x' mark by accident, but to conform this to what i said

spose we have to functions x(x) and y(x); or simply x and y

the derivative of xy would simply be x'y + xy'

x' = dx/dx which is just 1 so we tend to exclude it

x'y + xy' = y + xy'

Answer 19

ahhh

Answer 20

how did u get y+xy'??

Answer 21

i wrote it all out ... im not sure how much clearer i could have been with it.

so far we got

\[x(x+y)=x ^{4}y ^{2}-y\]
\[x^2+xy=x ^{4}y ^{2}-y\]
\[\frac d{dx}(x^2+xy=x ^{4}y ^{2}-y)\]
\[\frac d{dx}(x^2)+\frac d{dx}(xy)=\frac d{dx}(x ^{4}y ^{2})-\frac d{dx}(y)\]
\[\frac {dx}{dx}2x+\frac {dx}{dx}y+x\frac {dy}{dx}=\frac d{dx}(x ^{4}y ^{2})-\frac d{dx}(y)\]
\[2x+y+xy'=\frac d{dx}(x ^{4}y ^{2})-\frac d{dx}(y)\]

halfway there

Answer 22

the d/dx (y) at the end of it should seem easy enough

all thats left is the product of the powers

Answer 23

you might notice that we havent learned any new derivative rules ....

Answer 24

im trying to understand it

Answer 25

lets try some side stuff; if i ask you what the derivative of x^3 is, what would you say it is?

Answer 26

3x^2

Answer 27

but your almost right, i never said to derive it "with respect to x"

the derivative of x^3 is 3x^2 x' becasue of the chain rule

Answer 28

what is the derivative of 2y^2?

Answer 29

4y

Answer 30

dont forget the chain rule, without knwoing what the derivative is "with respect to" you have to pop out the prime of the variable

derivatove of 2y^2 = 4y y'

Answer 31

what is the derivative of 6t ?

Answer 32

6

Answer 33

if the derivative is y?? you have to put the y'?

Answer 34

your confusing a derivative with " a derivative with respect to __"

without knowing the "with respect to" variable we simply dont know that value of what the chain rule pops out

Answer 35

the derivative of y is: y'

the derivative of r is: r'

the derivative of p is: p'

the derivative of B is: B'

the derivative of 4g^3 is: 12g^2 g'

the derivative of x^2y is: 2xx' y + x^2 y'

Answer 36

you only begin to learn derivatives "with respect to x" but thats just to baby you into derivatives

Answer 37

any of this making sense yet?

Answer 38

if not, then youve simply learned derivatives the wrong way

Answer 39

not yet