Question

radical (xy)= 9 + x^2y find y' by implicit differentiation

Answer 1

wht do you get on your attempt?

Answer 2

i see chain rules and product rules for the most part ...

Answer 3

I think he can do the calculus part. I think the algebra is what scares him.

Answer 4

1/(2 radical x -4x radical x)

Answer 5

sqrt(xy)= 9 + x^2y

(x'y+xy')/(2sqrt(xy))= 2xx' y + x^2 y'

but x' wrt.x is dx/dx = 1

(y+xy')/(2sqrt(xy))= 2x y + x^2 y'

looking for more of the inner workings

Answer 6

to clean up . let 2 sqrt(xy) = u

y+xy'= 2ux y + ux^2 y'

xy'- ux^2 y'= 2ux y - y

y' (x- ux^2)= y(2ux - 1)

etc ...

Answer 7

d/dx radical(xy) = d/dx 9 + d/dx x^2 y
1/2 radical (xy) . y' = 0 + x^2 y' + 2xy

Answer 8

youve missed the chain rule on the left side

the derivative of (xy) pops out

Answer 9

recall:\[\sqrt{u}=u^{1/2}=1/2 u^{-1/2}~u'\]
but u =xy

therefore u' = x'y+xy' by the product rule

Answer 10

I don't get this part

Answer 11

derivative of radical(xy)

Answer 12

\[
\frac{d}{dx}\sqrt{xy} = \frac{d}{dx}(xy)^{1/2} = \frac 12 * (xy)^{-1/2} * \frac{d}{dx}(xy) = \\
\frac 12 * (xy)^{-1/2} * (xy' + y)
\]

Answer 13

you are aware of the chain rule, i can see that.

so to simplify it, let the innards of the radical be substituted by some function, u

what is the derivative of sqrt(u) wrt.x ?

Answer 14

oh i have to use the product rule to find the derivative of xy

Answer 15

correct

Answer 16

thanks

Answer 17

good luck :)

Answer 18

is it ( 4xy radical x - 4 ) / 2x^3 radical x