Question

Evaluate y' at the point (-1,2) for:

(3(x^3)y)-2(x^2)=8

Answer 1

I keep thinking I have it but keeps getting marked as wrong :/

Answer 2

Does it simplify to \[y'=\frac{ 4x-9x ^{2} y}{ 3x ^{3}y }\] ?

Answer 3

is the answer -2/9?

Answer 4

I don't know the answer. I got 11/3 but it said that was wrong

Answer 5

i just tried -2/9 also, that didn't work :/

Answer 6

okay well i put \[3x ^{3}y - 2x ^{2}\] is that right?

Answer 7

correct

Answer 8

\[3x^3y-2x^2\]

Answer 9

\[9x^2*y+3x^3*dy/dx -4x=0\]

Answer 10

okay well deriving the first part i got \[3x ^{3}\frac{ dy }{ dx } + 9x ^{2}y\] using product rule

Answer 11

\[3x^2*dy/dx=4x-9x^2*y\]

Answer 12

\[dy/dx =(4x-9x^2*y)/3x^2\]

Answer 13

\[(3(x^3)y)-2(x^2)=8 \rightarrow (3(x^3)y)=8+2(x^2)\]
\[y = \frac {8+2x^2}{3x^3}\]
differentiating w.r.t x we find:

\[y' = \frac {(3x^3) \frac{d}{dx}(8+2x^2) -(8+2x^2) \frac{d}{dx}(3x^3) }{(3x^3)^2}\]
\[y' = \frac {(3x^3)(0+2 \times 2x) -(8+2x^2) \(3 \times 3 x^2) }{(3x^3)^2}\]
\[y' = \frac {(3x^3)(4x) -(8+2x^2) (9 x^2) }{(3x^3)^2}\]
\[y' = \frac {12x^4 -(72x^2+18x^4)}{(3x^3)^2}\]
\[y' = \frac {12x^4 -72x^2-18x^4}{(3x^3)^2}\]
\[y' = \frac { -72x^2-6x^4}{9x^6}\]
\[y' = \frac { -6x^2(12+x^2)}{9x^6}\]
\[y' = \frac { -2(12+x^2)}{3x^4}\]
\[y'_{(-1, 2)} = \frac { -2(12+(-1)^2)}{3(-1)^4}= \frac { -2(12+1)}{3}\]
\[= \frac { -2(13)}{3}= \frac { -26}{3}\] is the required answer

@SRoyea

Answer 14

then you do the easy \[-4x = 0\]
solve for dy/dx
\[\frac{ dy }{ dx }=\frac{ 4x-9x ^{2}y }{ 3x ^{3} }\]

Answer 15

plug in and you get -14/3

Answer 16

Wow you guys are all really great thanks! Helped so much, got the answer and it makes sense to me now. The answer was 22/3. dpasingh, thanks for taking the time to break it down that far!

Answer 17

Thanks guys!!